补题：西南民族大学第十一届程序设计竞赛（同步赛）

比赛入口

B 都说小镇的切糕贵

做法：题意是要求一个字符串中能包含该串出现所有字符的最小长度，是滑动窗口技巧题目的典型例题。
具体滑动窗口实现：
如果左边界遇到与当前s[i]相同且s[i]之前出现过的，更新最小窗口长度res = min(res, i-l+1)；
如果左边界遇到与当前s[i]相同且s[i]之前未出现过的，把flag置为1，处理完l后强制更新res = i - l + 1。

代码：

#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {int f = 1; res = 0;char c = getchar();while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }res *= f;
}
int main() {fioint n; cin >> n;string s; cin >> s;int len = s.size();unordered_map<char, int> vis;vis.clear();int l = 0, res = n;for(int i = 0; i < len; ++i) {int flag = 0;if(!vis[s[i]]) flag = 1; //与当前s[i]相同且s[i]之前未出现过vis[s[i]]++;while(vis[s[l]] > 1) {vis[s[l]]--; l++;}if(flag) res = i - l + 1;else res = min(res, i - l + 1);}printf("%d\n", res);return 0;
}

C Guard the empire

做法：题意是要求最少需要多少个半径为r的圆（圆心都在x坐标上），能覆盖住n个给定坐标的点。做法就是找到每个圆能够取到的圆心位置的区间，再贪心找最少有多少个点能将所有区间都部分覆盖。

代码：

#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef pair<double, double> pdd;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {int f = 1; res = 0;char c = getchar();while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }res *= f;
}
const int N = 1003;
struct Qujian {int l, r;bool operator < (const Qujian &c) const {if(l != c.l) return l < c.l;return r < c.r;}
}qj[N];
double toQj(double y, double r) {return sqrt(r*r - y*y);
}
int main() {int n, c = 0;double r;while(scanf("%d%lf", &n, &r), n || r) {double x, y, cha;int flag = 1;for(int i = 0; i < n; ++i) {scanf("%lf%lf", &x, &y);if(y > r) {   //如果超出了圆能表示的范围flag = 0; continue;}cha = toQj(y, r);//找出可以覆盖住该点的圆心范围qj[i].l = x - cha; qj[i].r = x + cha;}if(!flag) {printf("Case %d: %d\n", ++c, -1);} else {//贪心找能覆盖住所有点的圆个数sort(qj, qj+n);double mid = qj[0].r;int res = 1;for(int i = 1; i < n; ++i) {if(qj[i].l > mid) {mid = qj[i].r; res++;} else if (qj[i].r < mid) {mid = qj[i].r;}}printf("Case %d: %d\n", ++c, res);}}return 0;
}

H Magic necklace

做法：有一串标号1~n的数字，要求每个数字不相邻的方案数，因为数字很小，直接上暴力预处理即可，这里用了next_permutation函数。

代码：

#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {int f = 1; res = 0;char c = getchar();while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }res *= f;
}
int res[15], mid[15], n;
int check(int n) {for(int i = 1; i <= n; ++i) {if(fabs(mid[i+1] - mid[i]) == 1) return 0;}return 1;
}
int main() {//暴力预处理res[1] = 1; res[2] = 0;for(int i = 3; i <= 11; ++i) {for(int j = 1; j <= i; ++j) mid[j] = j;mid[i+1] = mid[1];while(next_permutation(mid+2, mid+i+1)) {if(check(i)) res[i]++;}res[i] = res[i] / 2;}int t; scanf("%d", &t);while(t--) {scanf("%d", &n);printf("%d\n", res[n]);}return 0;
}

I 恋爱之子

做法：一些线段，所有相交的线段成为一个集合，要求有多少个集合，几何板子——判断两条线段是否相交，再用并查集即可。

代码：

#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {int f = 1; res = 0;char c = getchar();while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }res *= f;
}
const int N = 4003;
struct Dian {  //点LL x, y;
};
struct Xian {  //线Dian s, e;
} xian[N];
double mul(Dian a, Dian b, Dian c) {  //向量叉积return ((double)(b.x-a.x) * (c.y-a.y) - (double)(c.x-a.x) * (b.y-a.y));
}
int isJiao(Xian a, Xian b) {  //判断两个线段是否相交if( min(a.s.x, a.e.x) <= max(b.s.x, b.e.x) &&max(a.s.x, a.e.x) >= min(b.s.x, b.e.x) &&min(a.s.y, a.e.y) <= max(b.s.y, b.e.y) &&max(a.s.y, a.e.y) >= min(b.s.y, b.e.y) &&mul(b.s, a.s, b.e) * mul(b.s, b.e, a.e) >= 0 &&mul(a.s, b.s, a.e) * mul(a.s, a.e, b.e) >= 0 )return 1;return 0;
}
int a[N];
int Find(int x) {return a[x] == x ? x : a[x] = Find(a[x]);
}
void Merge(int x, int y) {int fx = Find(x), fy = Find(y);if(fx != fy) {a[fy] = fx;}
}
int main() {int n; scanf("%d", &n);for(int i = 0; i < n; ++i) {scanf("%lld%lld%lld%lld", &xian[i].s.x, &xian[i].s.y, &xian[i].e.x, &xian[i].e.y);a[i] = i;}for(int i = 0; i < n - 1; ++i) {for(int j = i + 1; j < n; ++j) {if(isJiao(xian[i], xian[j])) Merge(i, j);}}int res = 0;for(int i = 0; i < n; ++i) {if(a[i] == i) res++;}printf("%d\n", res);return 0;
}

K 集训队的依赖关系

做法：由一棵无根树建立超级源点0，再对这棵树进行树形dp，现在还是对树形dp很迷，等彻底学会再来写一篇树形dp专题，咕咕咕~

代码：

#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {int f = 1; res = 0;char c = getchar();while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }res *= f;
}
const int N = 5100;
int n, m, S;
int val[N], in[N], cost[N];
LL now[N], dp[N][N];
vector<int> ve[N];
void dfs(int u) {for(int i = S; i >= 0; --i) {if(i >= cost[u]) dp[u][i] = dp[u][i-cost[u]] + val[u];else dp[u][i] = -1e18;}for(int v : ve[u]) {for(int i = 0; i <= S; ++i) dp[v][i] = dp[u][i];dfs(v);for(int i = 0; i <= S; ++i) dp[u][i] = max(dp[u][i], dp[v][i]);}
}
int main() {int u, v;read(n); read(S); read(m);for(int i = 1; i <= n; ++i) read(val[i]);for(int i = 1; i <= n; ++i) read(cost[i]);for(int i = 0; i < m; ++i) {read(u); read(v);ve[v].push_back(u);in[u]++;}for(int i = 1; i <= n; ++i) {if(in[i]) continue;ve[0].push_back(i);}dfs(0);LL ans = 0;for(int i = 0; i <= S; ++i) ans = max(ans, dp[0][i]);printf("%lld\n", ans);return 0;
}

L 金牌厨师HiLin与HJGG

做法：题意就是有一个大矩阵，要求选一个最小的m，使得m*m矩阵中所有数字之和大于等于k，用到二位前缀和。

代码：

#include<bits/stdc++.h>
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define lson rt<<1, l, mid
#define rson rt<<1|1, mid+1, r
using namespace std;
typedef long long LL;
typedef pair<LL, int> pli;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef map<char, int> mci;
typedef map<string, int> msi;
template<class T>
void read(T &res) {int f = 1; res = 0;char c = getchar();while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }while(c >= '0' && c <= '9') { res = res * 10 + c - '0'; c = getchar(); }res *= f;
}
const int N = 2003;
int a[N][N];
LL sum[N][N];
int n, k;
int solve(int x, int y, int nn) {x--; y--;if(sum[x+nn][y+nn] - sum[x][y+nn] - sum[x+nn][y] + sum[x][y] >= k) return 1;return 0;
}
int main() {read(n); read(k);//二维前缀和预处理for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j) {read(a[i][j]);sum[i][j] = a[i][j] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];}}//暴力枚举每个m，加剪枝LL ans = n;for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j) {for(int nn = ans; nn >= 1 && nn <= n-i+1 && nn <= n-j+1; --nn) {if(solve(i, j, nn)) {ans = nn;} else break;}}}if(sum[n][n] < k) puts("I'm a Gold Chef!");else printf("%lld\n", ans);return 0;
}