将定期更新蓝桥杯习题集的解题报告~

ADV-251 Petri Net Simulation

cpp:

#include <algorithm>
#include <bitset>
#include <cassert>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <vector>using namespace std;const int maxn = 105;int p, t, np[maxn], lim;struct petri {int ip, op, i[maxn], o[maxn], in[maxn], out[maxn];
} pet[maxn];
// ip、op是参与变迁的place的编号int main() {int kase = 0;while (scanf("%d", &p) && p) {memset(pet, 0, sizeof(pet));for (int i = 1; i <= p; ++i) {scanf("%d", &np[i]);}scanf("%d", &t);for (int i = 1; i <= t; ++i) {int k;while (scanf("%d", &k), k) {if (k < 0) {++pet[i].in[-k];} else {++pet[i].out[k];}}for (int j = 1; j <= p; ++j) {if (pet[i].in[j]) {pet[i].i[++pet[i].ip] = j;}if (pet[i].out[j]) {pet[i].o[++pet[i].op] = j;}}}scanf("%d", &lim);int cnt = 0;for (int i = 1; i <= t; ++i) {bool flag = true;petri &k = pet[i];for (int j = 1; j <= k.ip; ++j) {if (np[k.i[j]] < k.in[k.i[j]]) {flag = false;break;}}if (!flag) continue;for (int j = 1; j <= k.ip; ++j) {np[k.i[j]] -= k.in[k.i[j]];}for (int j = 1; j <= k.op; ++j) {np[k.o[j]] += k.out[k.o[j]];}i = 0;if (++cnt >= lim) break;}if (cnt >= lim) {printf("Case %d: still live after %d transitions\n", ++kase, lim);} else {printf("Case %d: dead after %d transitions\n", ++kase, cnt);}printf("Places with tokens:");for (int i = 1; i <= p; ++i) {if (np[i]) {printf(" %d (%d)", i, np[i]);}}printf("\n\n");}return 0;
}

java:

import java.util.*;
public class Main {static int np, nt, nf;static int p[], in[][], out[][];static int cur;static int enabled() {for(int i=0; i<nt; i++) {boolean flag = true;for(int j=0; j<np; j++) if(p[j] < in[i][j]) {flag = false;break;}if(flag)return i;}return -1;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);int t = 0;while(true) {t++;cur = 0;np = sc.nextInt();if(np==0)break;p = new int[np];for(int i=0; i<np; i++) {p[i] = sc.nextInt();}nt = sc.nextInt();in = new int[nt][np];out = new int[nt][np];for(int i=0; i<nt; i++) {for(int j=0; ; j++) {int n = sc.nextInt();if(n==0)break;else if(n<0)in[i][-n-1] ++;elseout[i][n-1] ++;}}nf = sc.nextInt();while(true) {int n = enabled();if(n==-1||cur==nf)break; for(int j=0; j<np; j++) {p[j] -= in[n][j];p[j] += out[n][j];}cur++;} if(enabled()!=-1) {System.out.println("Case "+t+": still live after "+cur+" transitions");}else {System.out.println("Case "+t+": dead after "+cur+" transitions");}System.out.print("Places with tokens:");for(int i=0; i<np; i++) {if(p[i]>=1) System.out.print(" "+(i+1)+" ("+p[i]+")");}System.out.println("\n");}}
}

ADV-252 Navigation

cpp:

#include <math.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>struct SD {float x, y, r;
} sd[10];struct Dot {float x, y, d;bool flag;
} dot[2];int n;  //输出时的参数float distance(float x1, float y1, float x2, float y2) {return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}int belong(int x, int y) {if (fabs(sd[x].r - distance(sd[x].x, sd[x].y, dot[y].x, dot[y].y)) < 0.1)return 1;elsedot[y].flag = false;return 0;
}void find_dot(int x, int y) {float a, b, c, p, q, s, rx, ry, t[2];int i;rx = sd[x].x - sd[y].x;ry = sd[x].y - sd[y].y;a = 2 * sd[x].r * rx;b = 2 * sd[x].r * ry;c = sd[y].r * sd[y].r - sd[x].r * sd[x].r - rx * rx - ry * ry;p = a * a + b * b;q = -2 * a * c;s = c * c - b * b;t[0] = (sqrt(q * q - 4 * p * s) - q) / (2 * p);t[1] = (-sqrt(q * q - 4 * p * s) - q) / (2 * p);if (n == 2) {for (i = 0; i < 2; i++) {dot[i].d = asin(t[i]);dot[i].x = sd[x].x + sd[x].r * sin(dot[i].d);dot[i].y = sd[x].y + sd[x].r * cos(dot[i].d);if (belong(x, i) && belong(y, i)) dot[i].flag = true;}for (i = 0; i < 2; i++) {if (dot[i].flag == false) {dot[i].d = asin(t[i]);dot[i].x = sd[x].x + sd[x].r * sin(dot[i].d);dot[i].y = sd[x].y - sd[x].r * cos(dot[i].d);dot[i].flag = true;}}} else {dot[0].d = asin(t[0]);dot[0].x = sd[x].x + sd[x].r * sin(dot[0].d);dot[0].y = sd[x].y + sd[x].r * cos(dot[0].d);if (!(belong(x, 0) && belong(y, 0))) {dot[0].d = asin(-t[0]);dot[0].x = sd[x].x + sd[x].r * sin(dot[0].d);dot[0].y = sd[x].y + sd[x].r * cos(dot[0].d);}}
}int main() {int N;  //信号源数量int i, j;float tx, ty, rt;float t1 = -1, t2 = -1;float td;int d = 0;int k = 0;int a[10], b[10];do {n = -2;scanf("%d %f %f %f", &N, &rt, &tx, &ty);for (i = 0; i < N; i++) {float sx, sy, st, d;scanf("%f %f %f %f", &sx, &sy, &d, &st);sd[i].x = sx + 100.0 * sin(d * 3.1415926 / 180) * st;sd[i].y = sy + 100.0 * cos(d * 3.1415926 / 180) * st;sd[i].r = (rt - st) * 350.0;}dot[0].flag = false;dot[1].flag = false;if (N == 1) n = -1;for (i = 0; i < N; i++)for (j = i + 1; j < N; j++) {if ((distance(sd[i].x, sd[i].y, sd[j].x, sd[j].y) - sd[i].r -sd[j].r) < -0.1) {t1 = i;t2 = j;n = 2;break;}}if (n == -2) {for (i = 0; i < N; i++)for (j = i + 1; j < N; j++) {if (fabs(distance(sd[i].x, sd[i].y, sd[j].x, sd[j].y) -sd[i].r - sd[j].r) < 0.1) {t1 = i;t2 = j;n = 1;break;}}}if (n == 1 || n == 2) {find_dot(t1, t2);for (i = 0; i < N; i++)if (i != t1 && i != t2)for (j = 0; (j < 2) && (dot[j].flag == true); j++) {if (!belong(i, j)) {n--;}if (n == 0) break;}}if (n == 1) {for (i = 0; i < 2; i++) {if (dot[i].flag == true)if (distance(tx, ty, dot[i].x, dot[i].y) < 0.1)n = 3;else {td = atan(fabs((dot[i].y - ty) / (dot[i].x - tx))) *57.3;d = (int)(td + 0.5);if (tx > dot[i].x && ty < dot[i].y)d += 90;else if (tx < dot[i].x && ty < dot[i].y)d = 270 - d;else if (tx < dot[i].x && ty > dot[i].y)d += 270;else if (tx > dot[i].x && ty > dot[i].y)d = 90 - d;}}}a[k] = n;b[k++] = d;} while (N != 0);for (i = 0; i < k - 1; i++) {printf("Trial %d: ", i + 1);switch (a[i]) {case -1:printf("Inconclusive");break;case 0:printf("Inconsistent");break;case 1:printf("%d degrees", b[i]);break;case 2:printf("Inconclusive");break;case 3:printf("Arrived");}printf("\n");}return 0;
}

ADV-256 The Sky is the Limit

cpp:

#include <algorithm>
#include <complex>
#include <cstdio>
using namespace std;typedef long double db;
typedef complex<db> Point;
typedef Point Vector;int n, a[105], b[105], c[105], cn;
db cut[25005], rx, ans;db Cross(Vector A, Vector B) { return imag(conj(A) * B); }bool get_Cross(db px, db py, db qx, db qy, db p2x, db p2y, db q2x, db q2y) {Point P = (Point){px, py};Vector v = (Vector){qx - px, qy - py};Point Q = (Point){p2x, p2y};Vector w = (Vector){q2x - p2x, q2y - p2y};Vector u = P - Q;if (Cross(v, w) == 0) return false;db t = Cross(w, u) / Cross(v, w);rx = (P + v * t).real();if (rx < min(px, qx) || rx > max(px, qx)) return false;if (rx < min(p2x, q2x) || rx > max(p2x, q2x)) return false;return true;
}int main() {scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d%d%d", &a[i], &b[i], &c[i]);cut[++cn] = a[i] - c[i] / 2.0, cut[++cn] = a[i] + c[i] / 2.0,cut[++cn] = a[i];}for (int i = 1; i < n; i++)for (int j = i + 1; j <= n; j++) {db o = a[i] - c[i] / 2.0, p = a[i] + c[i] / 2.0,q = a[j] - c[j] / 2.0, r = a[j] + c[j] / 2.0;if (get_Cross(o, 0, a[i], b[i], q, 0, a[j], b[j])) cut[++cn] = rx;if (get_Cross(o, 0, a[i], b[i], a[j], b[j], r, 0)) cut[++cn] = rx;if (get_Cross(a[i], b[i], p, 0, q, 0, a[j], b[j])) cut[++cn] = rx;if (get_Cross(a[i], b[i], p, 0, a[j], b[j], r, 0)) cut[++cn] = rx;}sort(cut + 1, cut + cn + 1);cn = unique(cut + 1, cut + cn + 1) - cut - 1;for (int i = 1, pos; i < cn; i++) {pos = 0;db tmpl = -1E9, tmpr = -1E9, tl, tr;for (int j = 1; j <= n; j++)if (cut[i] >= a[j] - c[j] / 2.0 && cut[i + 1] <= a[j]) {tl = (cut[i] - a[j] + c[j] / 2.0) / (c[j] / 2.0) * b[j];tr = (cut[i + 1] - a[j] + c[j] / 2.0) / (c[j] / 2.0) * b[j];if (tl + tr >tmpl +tmpr)  //就是这个地方！不要写tl>tmpl||tr>tmpr，因为浮点误差的原因，有可能实际相同的点，反而低山峰的那个点更高那么一小丢丢，造成pos指向了低山峰。tmpl = tl, tmpr = tr, pos = j;} else if (cut[i] >= a[j] && cut[i + 1] <= a[j] + c[j] / 2.0) {tl = b[j] - (cut[i] - a[j]) / (c[j] / 2.0) * b[j];tr = b[j] - (cut[i + 1] - a[j]) / (c[j] / 2.0) * b[j];if (tl + tr > tmpl + tmpr) tmpl = tl, tmpr = tr, pos = j;}if (pos)ans += (cut[i + 1] - cut[i]) *sqrtl(1 +(b[pos] / (c[pos] / 2.0)) * (b[pos] / (c[pos] / 2.0)));}printf("%d", (int)(ans + 0.5));return 0;
}

java:

import java.util.*;
public class Main {public static void main(String[] args) {Scanner read = new Scanner(System.in);int N = read.nextInt();double num[][] = new double[N][3];List<Mon> liss = new ArrayList<Mon>();for (int i = 0; i < N; i++) {for (int j = 0; j < 3; j++)num[i][j] = read.nextDouble();}Line l[] = new Line[N];List<Node> list = new ArrayList<Node>();for (int i = 0; i < N; i++) {Line li = new Line();double x1 = num[i][0] - num[i][2] / 2;double x2 = num[i][0];double y2 = num[i][1];double x3 = num[i][0] + num[i][2] / 2;li.k1 = (y2 / (x2 - x1));       li.b1 = x1 * y2 / (x1 - x2);li.k2 = (y2 / (x2 - x3));li.b2 = x3 * y2 / (x3 - x2);l[i] = li;list.add(new Node(x1, 0));list.add(new Node(x2, y2));list.add(new Node(x3, 0));liss.add(new Mon(x2, y2, li));}Collections.sort(liss);for (int i = 0; i < N - 1; i++) {Mon n1 = liss.get(i);for (int j = i + 1; j < N; j++) {Mon n2 = liss.get(j);Line l1 = n1.l;Line l2 = n2.l;try {l1.x = (l2.b1 - l1.b1) / (l1.k1 - l2.k1);l1.y = l1.k1 * l1.x + l1.b1;if (l1.y >= 0 && l1.y <= n1.y && l1.y <= n2.y) {list.add(new Node(l1.x, l1.y));}} catch (Exception e) {}try {l1.x = (l2.b1 - l1.b2) / (l1.k2 - l2.k1);l1.y2 = l1.k2 * l1.x + l1.b2;if (l1.y2 >= 0 && l1.y2 <= n1.y && l1.y2 <= n2.y) {list.add(new Node(l1.x, l1.y2));}} catch (Exception e) {}try {l1.x = (l2.b2 - l1.b2) / (l1.k2 - l2.k2);l1.y2 = l1.k2 * l1.x + l1.b2;if (l1.y2 >= 0 && l1.y2 <= n1.y && l1.y2 <= n2.y) {list.add(new Node(l1.x, l1.y2));}} catch (Exception e) {}}}Collections.sort(list);for (int i = 0; i < list.size(); i++) {Node n = list.get(i);}double sum = 0;for (int i = 0; i < list.size() - 1; i++) {      Node n1 = list.get(i);Node n2 = list.get(i + 1);double x = (n2.x - n1.x) / 2 + n1.x;int ind = 0;int ind2 = 0;double max = 0;for (int j = 0; j < liss.size(); j++) {Mon m = liss.get(j);m.l.x = x;if (n2.x <= m.x) {m.l.y = m.l.k1 * m.l.x + m.l.b1;if (m.l.y > max) {ind = j;ind2 = 0;max = m.l.y;}} else {m.l.y2 = m.l.k2 * m.l.x + m.l.b2;if (m.l.y2 > max) {ind = j;ind2 = 1;max = m.l.y2;}}}Mon m = liss.get(ind);if (max == 0)continue;double y1 = 0, y2 = 0;if (ind2 == 0) {m.l.x = n1.x;m.l.y = m.l.k1 * m.l.x + m.l.b1;y1 = m.l.y;m.l.x = n2.x;m.l.y = m.l.k1 * m.l.x + m.l.b1;y2 = m.l.y;sum += Math.sqrt(Math.pow(n2.x - n1.x, 2) + Math.pow(y1 - y2, 2));} else {m.l.x = n1.x;m.l.y2 = m.l.k2 * m.l.x + m.l.b2;y1 = m.l.y2;m.l.x = n2.x;m.l.y2 = m.l.k2 * m.l.x + m.l.b2;y2 = m.l.y2;sum += Math.sqrt(Math.pow(n2.x - n1.x, 2) + Math.pow(y1 - y2, 2));}}System.out.println(String.format("%.0f", sum));}
}
class Mon implements Comparable<Mon> {Line l;double x;double y;public Mon(double x, double y, Line l) {super();this.x = x;this.y = y;this.l = l;}@Overridepublic int compareTo(Mon o) {// TODO Auto-generated method stubreturn (int) (this.x - o.x);}
}
class Line {double x;double k1, b1, k2, b2;double y;double y2;
}
class Node implements Comparable<Node> {double x;double y;public Node(double x, double y) {super();this.x = x;this.y = y;}@Overridepublic int compareTo(Node o) {// TODO Auto-generated method stubreturn (int) (this.x * 1000 - o.x * 1000);}
}

ADV-264 Degrees of Separation

cpp:

#include <stdio.h>
#include <algorithm>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <vector>
using namespace std;const int MAXN = 50 + 1;
vector<int> graph[MAXN];  //邻接表
int P, R;
bool vis[MAXN];  //记录结点是否访问
int dis[MAXN];   //到达结点的步数
map<string, int> People;
int ans = 0;void Bfs(int u) {  //标准的Bfs格式，因本题求最大分离度，略微复杂memset(dis, 0, sizeof(dis));memset(vis, 0, sizeof(vis));queue<int> Q;vis[u] = 1;Q.push(u);int num = 1;  //记录访问的结点个数while (!Q.empty()) {int now = Q.front();if (num == P &&Q.size() == 1)  //判断结点是否都访问过了，并且是最后一个结点ans = max(ans, dis[now]);Q.pop();for (int i = 0; i < graph[now].size(); i++) {int next = graph[now][i];if (!vis[next]) {vis[next] = 1;dis[next] = dis[now] + 1;  //步数加1num++;                     //已访问的结点加1Q.push(next);}}}
}int main() {int ccase = 0;while (scanf("%d%d", &P, &R) && P && R) {People.clear();for (int i = 0; i < MAXN; i++) graph[i].clear();ccase++;ans = 0;int k = 1;char c[20], d[20];// string c, d;for (int i = 0; i < R; i++) {// cin >> c >> d;scanf("%s%s", c, d);if (!People[c])  //将string映射到intPeople[c] = k++;if (!People[d]) People[d] = k++;graph[People[c]].push_back(People[d]);graph[People[d]].push_back(People[c]);}for (int i = 1; i <= P; i++) {Bfs(i);}if (ans == 0)printf("Network %d: DISCONNECTED\n\n", ccase);elseprintf("Network %d: %d\n\n", ccase, ans);}return 0;
}

java:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.StringTokenizer;
class R{static BufferedReader reader=new BufferedReader(new InputStreamReader(System.in));static StringTokenizer tokenizer=new StringTokenizer("");static String nextLine() throws IOException {return reader.readLine();}static String next() throws IOException {while(!tokenizer.hasMoreTokens()) {tokenizer=new StringTokenizer(reader.readLine());}return tokenizer.nextToken();}static int nextInt() throws NumberFormatException, IOException {return Integer.parseInt(next());}static Double nextDouble() throws NumberFormatException, IOException {return Double.parseDouble(next());}
}
public class Main {//static Scanner cin=new Scanner(System.in);static int p,r;static String s[]=new String[300];static int len[][]=new int[55][55];static String q;static Map mp=new HashMap<String, Integer>();public static void main(String[] args) throws IOException {// TODO Auto-generated method stubint cas=1;while(true) {p=R.nextInt();r=R.nextInt();if(p==0&&r==0) break;mp.clear();int cnt=1;for(int i=1;i<=p;i++) {for(int j=1;j<=p;j++) {len[i][j]=1000000000;}len[i][i]=0;}for(int i=0;i<r;i++) {String a,b;a=R.next();b=R.next();if(mp.containsKey(a)==false) {mp.put(a, cnt);cnt++;}if(mp.containsKey(b)==false) {mp.put(b, cnt);cnt++;}len[(int) mp.get(a)][(int) mp.get(b)]=1;len[(int) mp.get(b)][(int) mp.get(a)]=1;}int ans=0;for(int i=1;i<=p;i++) {for(int j=1;j<=p;j++) {for(int k=1;k<=p;k++) {len[j][k]=Math.min(len[j][k],len[j][i]+len[i][k]);}}}for(int i=1;i<=p;i++) {for(int j=1;j<=p;j++) {ans=Math.max(ans, len[i][j]);}}if(ans==1000000000) System.out.println("Network "+cas+": DISCONNECTED");else System.out.println("Network "+cas+": "+ans);System.out.println();cas++;}}
}

ADV-265 Cutting Chains

cpp:

#include <algorithm>
#include <cstring>
#include <iostream>
#define ms(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int maxn = 20 + 5;int G[maxn][maxn], vis[maxn], n, maxd, open[maxn], cnt = 0;bool dfs2(int p, int r) {for (int i = 1; i <= n; i++) {if (i == p || i == r) continue;if (G[p][i] && !open[i]) {if (!vis[i]) {vis[i] = 1;if (dfs2(i, p)) return true;} elsereturn true;}}return false;
}bool check() {for (int i = 1; i <= n; i++) {if (open[i]) continue;int s = 0;for (int j = 1; j <= n; j++) {if (G[i][j] && !open[j]) s++;}if (s > 2) return false;}ms(vis, 0);cnt = 0;for (int i = 1; i <= n; i++) {if (!open[i] && !vis[i]) {vis[i] = 1;if (dfs2(i, i)) return false;cnt++;}}return maxd >= cnt - 1;
}bool dfs(int p, int d) {if (d == maxd) {return check();}if (p > n) return false;open[p] = 1;if (dfs(p + 1, d + 1)) return true;open[p] = 0;if (dfs(p + 1, d)) return true;return false;
}int main() {int kase = 0;while (cin >> n && n) {int x, y;ms(G, 0);while (cin >> x >> y) {if (x == -1 && y == -1) break;G[x][y] = G[y][x] = 1;}int flag = 0;for (maxd = 0; !flag; maxd++) {ms(open, 0);flag = dfs(1, 0);}cout << "Set " << ++kase << ": Minimum links to open is ";cout << maxd - 1 << endl;}return 0;
}

java:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.Arrays;public class Main {static int INF =0x3f3f3f3f;static int vis[] = new int[20];static int g[][] = new int[20][20];static int n,  cn;static StreamTokenizer in;static void init() throws IOException {int a = 0, b = 0;g = new int[20][20];while (true) {in.nextToken();a=(int) in.nval;in.nextToken();b=(int) in.nval;if( a == -1 && b == -1)break;g[a - 1][b - 1] = g[b - 1][a - 1] = 1;}}static boolean two(int s) {for (int i = 0; i < n; i++) {if ((s&(1<<i))!=0) continue;int num = 0;for (int j = 0; j < n; j++) {if ((s&(1<<j))!=0) continue;if (g[i][j]!=0) num++;}if (num > 2) return true;}  return false;}static boolean dfs(int s, int now, int fa) {vis[now] = 1;for (int i = 0; i < n; i++) {if (!(g[now][i]!=0) || (s&(1<<i))!=0 || i == fa) continue;if (vis[i]!=0) return true;if (dfs(s, i, now)) return true;}return false;}static boolean circle(int s) {for (int i = 0; i < n; i++) {if (vis[i]!=0|| (s&(1<<i))!=0) continue;cn++;if (dfs(s, i, -1)) return true;}return false;}static int cal(int s) {return s == 0 ? 0 : cal(s / 2) + (s&1);}static int solve() {int ans = INF;int s = (1<<n);for (int i = 0; i < s; i++) {cn = 0; vis = new int[20];if(two(i) || circle(i)) continue;if (cal(i) >= cn - 1)ans = Math.min(cal(i), ans);}return ans;}public static void main(String[] args) throws IOException {in= new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));int cas = 0;while (true) {in.nextToken();n=(int) in.nval;if(n==0)break;init();System.out.println("Set "+ ++cas+": Minimum links to open is "+solve());}}}

ADV-270 Flight Planning

cpp:

#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
using namespace std;
const int maxn = 1000 + 10;
long long dp[maxn][maxn];
int ans[maxn], s[maxn];
int he[maxn][maxn];
int main() {int x;scanf("%d", &x);int u = 1;while (x--) {int a, b, c, n;scanf("%d", &n);scanf("%d %d %d", &a, &b, &c);for (int i = 20; i <= 40; i++) {double w = (c - b) * (i - 20) * 1.0 / 20.0 + b;double v = 400 + w;double t = a / (1.0 * v);dp[1][i] = ((fabs(i - 30) * 10 + 2000) * t + i * 50) * 10000;}for (int i = 2; i <= n; i++) {scanf("%d %d %d", &a, &b, &c);for (int j = 20; j <= 40; j++) {dp[i][j] = LONG_LONG_MAX;double w = (c - b) * (j - 20) * 1.0 / 20.0 + b;double v = 400 + w;double t = a / (1.0 * v);for (int k = 20; k <= 40; k++) {long long o =dp[i - 1][k] + (fabs(j - 30) * 10 + 2000) * t * 10000;if (j > k) o += (j - k) * 500000;if (dp[i][j] > o) {he[i][j] = k;dp[i][j] = o;}}}}for (int i = 1; i <= n; i++) ans[i] = INT_MAX;long long mi = LONG_LONG_MAX;for (int i = 20; i <= 40; i++) {mi = min(mi, dp[n][i]);}for (int i = 20; i <= 40; i++) {if (dp[n][i] == mi) {int p = n, c = i;while (he[p][c]) {s[p] = c;c = he[p][c];p--;}s[p] = c;int flag = 0;for (int j = 1; j <= n; j++) {if (s[j] < ans[j]) {flag++;break;}if (s[j] > ans[j]) {break;}}if (flag) {for (int j = 1; j <= n; j++) {ans[j] = s[j];}}}}printf("Flight %d: ", u++);for (int i = 1; i <= n; i++) {printf("%d ", ans[i]);}printf("%lld\n", mi / 10000 + 1);}return 0;
}

java:

import java.util.*;
public class Main {static Scanner sc=new Scanner(System.in);static int V=400;static int GP=2000,GE=10,C=50;static int A=30;   static double dp[][];static int m[][];      //记忆前一段的高度static int Arr[][];   //A[i][0]记录最低点风速 A[i][1]记录最高点风速static int H[];          //长度数组static int K;         //个数static void init() {        //没问题K=sc.nextInt();H=new int[K+1];dp=new double[K+1][21];m=new int[K+1][21];Arr=new int[K+1][2];for(int i=1;i<=K;i++) {H[i]=sc.nextInt();Arr[i][0]=sc.nextInt();Arr[i][1]=sc.nextInt();}for(int j=0;j<=20;j++) {dp[0][j]=C*(j+20);}}public static void main(String[] args) {int n=sc.nextInt();for(int i=1;i<=n;i++) {init();for(int k=1;k<=K;k++)for(int j=0;j<=20;j++)getDP(k,j);double min=1000000000;int p=-1;for(int j=0;j<=20;j++) {if(dp[K][j]<min) {min=dp[K][j];p=j;}}print(p,i);System.out.println((int)Math.ceil(min));}}static void print(int k,int j) {int arr[]=new int[K];int p=k;for(int i=K;i>=1;i--) {arr[i-1]=k=m[i][k];}System.out.print("Flight "+j+": ");for(int i=1;i<K;i++) {System.out.print(arr[i]+20+" ");}System.out.print((p+20)+" ");}static void getDP(int i,int j) {  //double min=100000000;for(int k=0;k<=20;k++) {double num=dp[i-1][k];if(k<j)num+=50*(j-k);if(min>num) {m[i][j]=k;min=num;}}min+=getHY(i,j);dp[i][j]=min;//System.out.println(i+" "+j+" "+dp[i][j]);}static double getHY(int i,int h) {    //第i段航线高度为h的耗油数 没问题//System.out.println("第"+i+"段，风速为："+(V+getV(i,h))+" 高度为："+h+"  耗油数为："+getY(h)+" 总耗油为："+(H[i]*getY(h)/(V+getV(i,h))));return H[i]*getY(h)/(V+getV(i,h));}static double getY(int h) {return 2000+Math.abs(10-h)*10;}static double getV(int i,int x) {        //得到第i段航路高度为x的风速return (double)(Arr[i][1]-Arr[i][0])*x/20+Arr[i][0];}
}

ADV-271 Stacking Plates

cpp:

#include <algorithm>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <set>
#include <vector>using namespace std;const int INF = 0x3f3f3f3f;int kase, n, id[10005], vis[55][2505], dp[2505][55];
// id[i]为盘子直径i离散化后的值, vis[i][j]判断堆i中是否出现过第j种盘子,
// dp[i][j]表示把前i种盘子排好放到堆j的最小花费
vector<int> plates[55];  //输入数据
set<int> val;            //用于离散化
vector<int> heap[2505];  // heap[i]表示第i种盘子对应的堆// 初始化全局变量
void init() {memset(dp, INF, sizeof(dp));memset(vis, false, sizeof(vis));for (int i = 1; i <= n; i++) plates[i].clear();val.clear();for (int i = 1; i <= 2500; i++) heap[i].clear();
}// 离散化并返回不同直径盘子的种类
int discrete() {int ran = 0;for (set<int>::iterator it = val.begin(); it != val.end(); it++)id[*it] = ++ran;return ran;
}int main() {while (~scanf("%d", &n)) {init();// 处理输入数据int k, x;for (int i = 1; i <= n; i++) {scanf("%d", &k);while (k--) {scanf("%d", &x);plates[i].push_back(x);val.insert(x);}}// 离散化以及算出每个类型盘子所在的堆idint ran = discrete();for (int i = 1; i <= n; i++) {for (int j = 0; j < plates[i].size(); j++) {x = id[plates[i][j]];if (!vis[i][x]) {vis[i][x] = 1;heap[x].push_back(i);  //保证第x种盘子对应的堆不重复}}}// 动态规划算出最优值for (int i = 0; i < heap[1].size(); i++)dp[1][heap[1][i]] = heap[1].size() - 1;for (int i = 2; i <= ran; i++) {for (int a = 0; a < heap[i].size(); a++) {for (int b = 0; b < heap[i - 1].size(); b++) {int j = heap[i][a], k = heap[i - 1][b],cnt = heap[i].size();if (j == k)dp[i][j] = min(dp[i][j],dp[i - 1][j] +(cnt == 1 ? 0 : cnt));  // cnt-1==0?0:cnt-1+1elsedp[i][j] =min(dp[i][j], dp[i - 1][k] + cnt - 1 + !vis[k][i]);}}}// 处理及输出结果int ans = INF;for (int i = 0; i < heap[ran].size(); i++)ans = min(ans, dp[ran][heap[ran][i]]);printf("Case %d: %d\n", ++kase, 2 * ans - n + 1);}return 0;
}

ADV-275 JOE的算数

cpp:

#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;
inline int read() {int f = 1, x = 0;char ch = getchar();while (ch < '0' || ch > '9') {if (ch == '-') f = -1;ch = getchar();}while (ch >= '0' && ch <= '9') {x = x * 10 + ch - 48;ch = getchar();}return f * x;
}
ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll poww(ll x, ll y, ll p) {ll ans = 1;while (y) {if (y & 1) ans = ans * x % p;x = x * x % p;y >>= 1;}return ans;
}
ll a, b, c;
int main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);  // open-> can't use scanf,printf!!cin >> a >> b >> c;cout << poww(a, b, c) << endl;return 0;
}

java:

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);long a=sc.nextLong();int b=sc.nextInt();int c=sc.nextInt();long res=1;while (b!=0) {if ((b&1)==1) {res=(res*a)%c;}a=(a*a)%c;b>>=1;}System.out.println(res);}
}

ADV-276 Raising the Roof

暂时未通过

ADV-277 The Islands

cpp:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 104;
struct point {int x, y;point() {}point(int xx, int yy) {this->x = xx;this->y = yy;}
} a[maxn], pre[maxn][maxn];int inline p2(int x) { return x * x; }
inline double dis(point a, point b) {return sqrt(p2(a.x - b.x) + p2(a.y - b.y));
}int n, m1, m2;
double dp[maxn][maxn];
bool vis[maxn][maxn];
int main() {int cas = 1;while (scanf("%d%d%d", &n, &m1, &m2) != EOF) {if (n == 0 && m1 == 0 && m2 == 0) {break;}m1++, m2++;for (int i = 1; i <= n; i++) {scanf("%d%d", &a[i].x, &a[i].y);}for (int i = 0; i <= n; i++) {for (int j = 0; j <= n; j++) {vis[i][j] = false;dp[i][j] = 1e10;}}dp[1][1] = 0;for (int i = 1; i < n; i++) {for (int j = 1; j < n; j++) {if (i != 1 && i == j) continue;if (i >= j) {  // i+1的去哪// dp[i][j] -> dp[i+1][j] = dp[i][j]+dis(i,i+1)  dp[i][i+1]// = dp[i][j]+dis(j,i+1)if (i + 1 != m2 &&(vis[i + 1][j] == false ||dp[i + 1][j] > dp[i][j] + dis(a[i], a[i + 1]))) {pre[i + 1][j] = point(i, j);vis[i + 1][j] = true;dp[i + 1][j] = dp[i][j] + dis(a[i], a[i + 1]);// printf("%d %d -> %d %d (1)\n",i,j,i+1,j);}if (i + 1 != m1 &&(vis[i][i + 1] == false ||dp[i][i + 1] > dp[i][j] + dis(a[j], a[i + 1]))) {pre[i][i + 1] = point(i, j);vis[i][i + 1] = true;dp[i][i + 1] = dp[i][j] + dis(a[j], a[i + 1]);// printf("%d %d -> %d %d (2)\n",i,j,i,i+1);}}if (i <= j) {  // j+1的去哪// dp[i][j] -> dp[i][j+1] = dp[i][j]+dis(j,j+1)  dp[j+1][j]// = dp[i][j]+dis(i,j+1)if (j + 1 != m1 &&(vis[i][j + 1] == false ||dp[i][j + 1] > dp[i][j] + dis(a[j], a[j + 1]))) {pre[i][j + 1] = point(i, j);vis[i][j + 1] = true;dp[i][j + 1] = dp[i][j] + dis(a[j], a[j + 1]);// printf("%d %d -> %d %d (3)\n",i,j,i,j+1);}if (j + 1 != m2 &&(vis[j + 1][j] == false ||dp[j + 1][j] > dp[i][j] + dis(a[i], a[j + 1]))) {pre[j + 1][j] = point(i, j);vis[j + 1][j] = true;dp[j + 1][j] = dp[i][j] + dis(a[i], a[j + 1]);// printf("%d %d -> %d %d (4)\n",i,j,j+1,j);}}}}/*for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){printf("dp[%d][%d] = %15.2f  ",i,j,dp[i][j]);}printf("\n");}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){printf("pre[%d][%d] = (%d, %d)  ",i,j,pre[i][j].x, pre[i][j].y);}printf("\n");}*/double ans = 1e20;int posx = -1, posy = n;for (int i = 1; i <= n; i++) {if (i == m2) continue;if (ans > dp[i][n] + dis(a[i], a[n])) {ans = dp[i][n] + dis(a[i], a[n]);posx = i;}}vector<int> way1, way2;while (posx != 0 && posy != 0) {// printf("%d %d ->",posx,posy);if (pre[posx][posy].x != posx) {way1.push_back(posx);posx = pre[posx][posy].x;} else {way2.push_back(posy);posy = pre[posx][posy].y;}}vector<int> xcx;printf("Case %d: %.2f\n", cas++, ans);for (int i = way1.size() - 1; i >= 0; i--) {xcx.push_back(way1[i] - 1);}for (int i = 0; i < way2.size(); i++) {xcx.push_back(way2[i] - 1);}xcx.push_back(0);if (xcx[1] != 1) {reverse(xcx.begin(), xcx.end());}for (int i = 0; i < xcx.size(); i++) {if (i != 0) {printf(" ");}printf("%d", xcx[i]);}printf("\n");}return 0;
}

ADV-278 歌唱比赛

cpp:

#include <stdio.h>
int main() {int n;scanf("%d", &n);while (n--) {double sum = 0;int a, b, c, d;scanf("%d%d%d%d", &a, &b, &c, &d);sum += a * 0.7 + b * 0.2 + c * 0.1 + d;sum = sum > 100 ? 100 : sum;printf("%0.1f\n", sum);}
}

java:

import java.text.DecimalFormat;
import java.util.*;
public class Main {/*** @param args*/public static void main(String[] args) {// TODO Auto-generated method stubScanner reader = new Scanner(System.in);DecimalFormat df = new DecimalFormat("#.0");int n;n = reader.nextInt();int a,b,c,d;double[] sum = new double[n];for(int i=0;i<n;i++){a = reader.nextInt();b = reader.nextInt();c = reader.nextInt();d = reader.nextInt();sum[i] = 0.7*a+0.2*b+0.1*c+d;if(sum[i] > 100)sum[i] = 100; }for(int i=0;i<n;i++)System.out.print(df.format(sum[i])+"\n");//System.out.print(String.format("%.1f", sum));    }}

ADV-279 矩阵乘法

cpp:

#include <stdio.h>
int main() {int n, m, k;scanf("%d%d%d", &n, &m, &k);int a[n + 1][m + 1], b[m + 1][k + 1];for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++) scanf("%d", &a[i][j]);for (int i = 1; i <= m; i++)for (int j = 1; j <= k; j++) scanf("%d", &b[i][j]);int c[n + 1][k + 1];int x = 1;for (int i = 1; i <= n; i++)for (int j = 1; j <= k; j++) {x = 1;c[i][j] = 0;while (x <= m) {c[i][j] += a[i][x] * b[x][j];x++;}}for (int i = 1; i <= n; i++) {for (int j = 1; j <= k; j++) printf("%d ", c[i][j]);printf("\n");}return 0;
}

java:

import java.util.Scanner;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);int n = sc.nextInt();int m = sc.nextInt();int k = sc.nextInt();int mar1[][] = new int[n][m];int mar2[][] = new int[m][k];for(int i=0;i<n;i++) {for(int j=0;j<m;j++) {mar1[i][j] = sc.nextInt();}}for(int i=0;i<m;i++) {for(int j=0;j<k;j++) {mar2[i][j] = sc.nextInt();}}int ans[][] = new int[n][k];for(int i=0;i<n;i++) {for(int j=0;j<k;j++) {for(int t=0;t<m;t++) {ans[i][j] += mar1[i][t]*mar2[t][j];}System.out.print(ans[i][j]+" ");}System.out.println();}}
}

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