ACM-ICPC 2018 北京赛区网络预赛（A B C D H）

自闭症患者

A. Saving Tang Monk II

题目链接

题面：

划掉

题意：

大搜索，队友写的

思路：

大搜索，队友写的

AC代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
char pic[102][102];
int dp[6][102][102];
struct step {int x, y, w;step() {}step(int _x, int _y, int _w) : x(_x), y(_y), w(_w) {}
};
queue<step> bfs;
int dirx[4] = { -1, 0, 1, 0 };
int diry[4] = { 0, 1, 0, -1 };
int main() {int x, y;while (scanf("%d %d", &x, &y) != EOF && x > 0 && y > 0) {for (int i = 0; i < x; i++) {scanf("%s", pic[i]);}memset(dp, 0x3f3f3f3f, sizeof(dp));while (!bfs.empty()) bfs.pop();int sx, sy, tx, ty;for (int i = 0; i < x; i++) {for (int j = 0; j < y; j++) {if (pic[i][j] == 'S') sx = i, sy = j;if (pic[i][j] == 'T') tx = i, ty = j;}}dp[0][sx][sy] = 0;bfs.push(step(sx, sy, 0));while (!bfs.empty()) {step topx = bfs.front();bfs.pop();for (int k = 0; k < 4; k++) {int nt = dp[topx.w][topx.x][topx.y];int nx = topx.x + dirx[k];int ny = topx.y + diry[k];int nw = topx.w;if (nx >= 0 && ny >= 0 && nx < x && ny < y) {if (pic[nx][ny] == '#') {if (nw == 0) continue;nw--;nt++;} else if (pic[nx][ny] == 'B') {if (nw < 5) nw++;} if (pic[topx.x][topx.y] != 'P') {nt++;}if (nt < dp[nw][nx][ny]) {dp[nw][nx][ny] = nt;if (pic[nx][ny] != 'T') {bfs.push(step(nx, ny, nw));}}}}}
//         for (int w = 0; w <= 5; w++) {//             for (int i = 0; i < x; i++) {//                 for (int j = 0; j < y; j++) {//                     printf("%d ", dp[w][i][j]);
//                 }printf("\n");
//             }printf("\n\n");
//         }int ans = 0x3f3f3f3f;for (int i = 0; i <= 5; i++) {ans = min(ans, dp[i][tx][ty]);}printf("%d\n", ans == 0x3f3f3f3f ? -1 : ans);}return 0;
}

B. Tomb Raider

题目链接

题面：

划掉

题意：

寻址n个串的最长公共子序列。

思路：

拿一个最短串枚举所有子序列，在其他串中查找。暴力

AC代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <cmath>
#include <map>using namespace std;typedef long long LL;const int MAXN = 1000005;string s[12];
int use[10];
string ans;int main()
{int n;string temp;string now;while(cin >> n){for(int i = 0; i < n; ++i){cin >> s[i];}for(int i = 1; i < n; ++i){s[i] += s[i];//cout << s[i] << endl;}int len = s[0].length();int maxx = (1<<len);ans = "";for(int i = 0; i < maxx; ++i){temp = "";int cnt = 0;for(int j = 0; j < len; ++j){if(i&(1<<j)){temp += s[0][j];}}int tl = temp.length();for(int x = 0; x < tl; ++x){now = temp.substr(x+1, tl-(x+1)) + temp.substr(0, x);now.push_back(temp[x]);int nl = now.length();//cout << now << endl;bool flag = true;bool tempflag;for(int j = 1; j < n; ++j){int lj = s[j].length()/2;tempflag = false;for(int st = 0; st < lj; ++st){if(s[j][st] != now[0]) continue;int cur = 1;if(nl == 1){tempflag = true;break;}for(int k = st+1; k < st+lj; ++k){if(s[j][k] == now[cur]){++cur;if(cur == nl){tempflag = true;break;}}}if(tempflag) break;}if(!tempflag){flag = false;break;}}if(flag){//cout << now << endl;if(now.length() > ans.length()){ans = now;reverse(now.begin(), now.end());if(now < ans){ans = now;}}else if(now.length() == ans.length()){if(now < ans){ans = now;}reverse(now.begin(), now.end());if(now < ans){ans = now;}}}}}if(ans == ""){cout << 0 << endl;}else{cout << ans << endl;}}
}

C. Cheat

题目链接

题面：

划掉

题意：

斗地主。

思路：

大模拟，队友写的。

AC代码：

#include <cmath>
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
int dictOrder[] = {9, 1, 2, 3, 4, 5, 6, 7, 8, 0, 10, 12, 11};
int cards[4][13];
int fuck[4][13];
int tables[13];
char fv[5];
int rankNow = 0, rounderRank = -1, rounderCnt = -1, roundext = -1, roundNow = 0;
int getcard(char* sp) {#define card(a, b) if (strcmp(sp, a) == 0) return b;card("A", 0); card("2", 1); card("3", 2); card("4", 3); card("5", 4); card("6", 5);card("7", 6); card("8", 7); card("9", 8); card("10", 9); card("J", 10); card("Q", 11);card("K", 12);return -1;
#undef card
}
const char* getcardII(int sp) {#define card(b, a) if (sp == a) return b;card("A", 0); card("2", 1); card("3", 2); card("4", 3); card("5", 4); card("6", 5);card("7", 6); card("8", 7); card("9", 8); card("10", 9); card("J", 10); card("Q", 11);card("K", 12);return "?";
#undef card
}
int findSmallest(int p) {for (int i = 0; i < 13; i++) {if (cards[p][dictOrder[i]] > 0) {return dictOrder[i];}}return -1;
}
int findNextCard(int p) {for (int i = 0; i < 13; i++) {if (fuck[p][i] > 0) {fuck[p][i]--;return i;}}return -1;
}
void putOneDictSmallest(int p) {int ck = findSmallest(p);if (ck >= 0) {cards[p][ck]--;rounderCnt = 1;rounderRank = ck;tables[ck]++;}
}
void putSomeTrueCard(int p, int cnt) {rounderCnt = cnt;cards[p][rankNow] -= cnt;rounderRank = rankNow;tables[rankNow] += cnt;
}
bool action(int p, bool realDo, int rankx) {if (realDo) roundext = -1;if (p == 0) {if (cards[p][rankx] > 0) {if (realDo) putSomeTrueCard(p, 1);return true;} else {if (realDo) putOneDictSmallest(p);return false;}} else if (p == 1) {if (cards[p][rankx] > 0) {if (realDo) putSomeTrueCard(p, cards[p][rankx]);return true;} else {if (realDo) putOneDictSmallest(p);return false;}} else if (p == 2) {if (cards[p][rankx] > 0) {if (realDo) putSomeTrueCard(p, cards[p][rankx]);return true;} else {if (realDo) {int nums = 9999, idx = -1;for (int i = 0; i < 13; i++) {if (cards[p][dictOrder[i]] > 0) {if (cards[p][dictOrder[i]] < nums) {nums = cards[p][dictOrder[i]];idx = dictOrder[i];}}}rounderCnt = cards[p][idx];cards[p][idx] = 0;rounderRank = idx;tables[idx] += rounderCnt;}return false;}} else if (p == 3) {if (cards[p][rankx] > 2) {if (realDo) putSomeTrueCard(p, cards[p][rankx]);return true;} else {if (realDo) {putSomeTrueCard(p, cards[p][rankx]);roundext = findSmallest(p);if (roundext >= 0) {cards[p][roundext]--;tables[roundext]++;}return roundext == -1;}return false;}}
}
bool shouldFuck(int p) {if (p == 0) {if ((roundNow + 1) % 4 == p && !action(p, false, (rankNow + 1) % 13)) {return true;} else {return cards[p][rankNow] + (rounderCnt + (roundext >= 0)) > 4;}} else if (p == 1) {return (roundNow + 1) % 4 == p && !action(p, false, (rankNow + 1) % 13);} else if (p == 2) {return cards[p][rankNow] == 4;} else if (p == 3) {return findSmallest(roundNow % 4) == -1;}
}
void getAllCards(int p) {for (int i = 0; i < 13; i++) {cards[p][i] += tables[i];tables[i] = 0;}
}
void printCards() {memcpy(fuck, cards, sizeof(cards));for (int i = 0; i < 4; i++) {for (int k = 0; k < 998; k++) {int next = findNextCard(i);if (next == -1) {if (k == 0) printf("WINNER");printf("\n");break;}else {if (k > 0) printf(" ");printf("%s", getcardII(next));}}}
}
int main() {bool gaming = true;while (gaming) {memset(cards, 0, sizeof(cards));memset(tables, 0, sizeof(tables));for (int i = 0, j = -1; i < 52; i++) {if (i % 13 == 0) j++;if (scanf("%s", fv) == EOF) {gaming = false;break;}int cc = getcard(fv);if (cc >= 0) cards[j][cc]++;}rankNow = 0, roundNow = 0, rounderRank = -1, rounderCnt = -1, roundext = -1;if (gaming) {while (true) {//printf("round: %s\n", getcardII(rankNow));bool really = action(roundNow % 4, true, rankNow);//printCards();//printf("rankcnt %d ext: %s\n", rounderCnt, getcardII(roundext));//printf("p%d is really: %d\n", roundNow % 4, really);for (int j = roundNow + 1; j < roundNow + 4; j++) {if (shouldFuck(j % 4)) {//printf("p%d do fuck!\n", j % 4);if (!really) {getAllCards(roundNow % 4);} else {getAllCards(j % 4);}break;}}rankNow = (rankNow + 1) % 13;if (findSmallest(roundNow % 4) == -1) break;roundNow++;}printCards();}}return 0;
}

D. 80Days

题目链接

题面：

划掉

题意：

n个成环的城市，到达第i个城市获得aia_iai，离开丢掉bib_ibi，初始有c，全程拥有值不能为负数，问是否可行，可行就输出最小的出发点。

思路：

尺取法，维护一个连续段的区间以及C+在该区间旅游费用和，左端点不变，右端点向右移，如果遇到无法进行的情况则移动左端点。

AC代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#include <cmath>
#include <map>using namespace std;typedef long long LL;const int MAXN = 1000005;LL a[2*MAXN];
LL b[2*MAXN];int main()
{int t;cin >> t;int n;LL c;while(t--){scanf("%d%lld", &n, &c);for(int i = 1; i <= n; ++i){scanf("%lld", &a[i]);a[n+i] = a[i];}for(int i = 1; i <= n; ++i){scanf("%lld", &b[i]);b[n+i] = b[i];}int l = 1, r = 1;LL now = c + a[1];bool flag = false;while(true){now -= b[r];while(now < 0 && l < r){now -= a[l];now += b[l];l++;}if(l == r && now < 0){l++;r++;now = c + a[l];continue;}r++;if(r-l == n){flag = true;break;}now += a[r];while(now < 0 && l < r){now -= a[l];now += b[l];l++;}if(l > n){flag = false;break;}}if(flag){printf("%d\n", l);}else{printf("-1\n");}}
}

H. K-Demensional Foil II

题目链接

题面：

划掉

题意：

K维空间，给定一个曼哈顿距离意义下的球，N个点，求这些点到该球上在欧几里得距离下最小的点。

思路：

直接考虑对欧几里得距离进行缩小，那么可以写成一个类似于方差形式的表达式，那么要使得该表达式最小，令这K维当中对表达式值贡献最大的值减小。
问题变形为满足哈密顿距离小于R时放缩该表达式使得其值最小，二分即可。

AC代码：

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <stack>
#include <bitset>using namespace std;#define FSIO  ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define DEBUG(a)   cout<<"DEBUG: "<<(a)<<endl;
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define X  first
#define Y  second
#define REP(i,st,ed)    for(int i=st;i<=ed;++i)
#define IREP(i,st,ed)   for(int i=ed;i>=st;--i)
#define TCASE(T)    cin>>T;while(T--)const int MAXN = 105;
const int MOD = 2333333 ;
const int INF = 1e9+7;int _;using namespace std;const double EPS = 1e-6;double foil[MAXN];
double spcf[MAXN];
double go[MAXN];
bool mk[MAXN];
int N, K;
double R;int main()
{//freopen("test.in","r",stdin);//freopen("test.out","w+",stdin);TCASE(_){scanf("%d%d",&N,&K);scanf("%lf",&R);REP(i,1,K)  scanf("%lf",&foil[i]);REP(t,1,N){REP(i,1,K){scanf("%lf",&spcf[i]);go[i] = fabs(spcf[i]-foil[i]);mk[i] = (spcf[i]>=foil[i]);}double minn = 0;double maxn = 1e8;double mid, tmp;while(minn+EPS<=maxn){mid = (minn+maxn)/2.0;tmp = 0;REP(kk,1,K) tmp+=max(0.0,go[kk]-mid);if(tmp<=R)  maxn = mid-EPS;else    minn = mid+EPS;}REP(i,1,K){if(go[i]>=maxn)    go[i]=maxn;if(mk[i])   tmp = spcf[i]-go[i];else    tmp = spcf[i]+go[i];printf("%.5f ",tmp);}printf("\n");}}return 0;
}

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ACM-ICPC 2018 北京赛区网络预赛（A B C D H）

ACM-ICPC 2018 北京赛区网络预赛（A B C D H）

A. Saving Tang Monk II

题目链接

题面：

题意：

思路：

AC代码：

B. Tomb Raider

题目链接

题面：

题意：

思路：

AC代码：

C. Cheat

题目链接

题面：

题意：

思路：

AC代码：

D. 80Days

题目链接

题面：

题意：

思路：

AC代码：

H. K-Demensional Foil II

题目链接

题面：

题意：

思路：

AC代码：

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