最后dp[ (1<<n) -1 ][ m ]表示每行每列各取一个数，最后取n个数并得分大于等于m的方案数。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;int dp[1<<12][510];
int f[13];
int a[13][13];int gcd(int a, int b)
{if(b == 0) return a;return gcd(b,a%b);
}int main()
{int test;int n,m;f[0] = 1;for(int i = 1; i <= 12; i++)f[i] = f[i-1] * i;scanf("%d",&test);while(test--){scanf("%d %d",&n,&m);for(int i = 1; i <= n; i++)for(int j = 1; j <= n; j++)scanf("%d",&a[i][j]);for(int i = 0; i < (1<<n); i++)for(int j = 0; j <= m; j++)dp[i][j] = 0;dp[0][0] = 1;for(int i = 0; i < (1<<n); i++){int cnt = 0;for(int j = 1; j <= n; j++){if(i & (1<<(j-1)) )cnt++;}for(int j = 1; j <= n; j++){if(i & (1<<(j-1))) continue;for(int g = 0; g <= m; g++){if(g + a[cnt+1][j] >= m)dp[i+(1<<(j-1))][m] += dp[i][g];elsedp[i+(1<<(j-1))][g+a[cnt+1][j]] += dp[i][g];}}}if(dp[(1<<n)-1][m] == 0)printf("No solution\n");else{int tmp = gcd(f[n],dp[(1<<n)-1][m]);printf("%d/%d\n",f[n]/tmp, dp[(1<<n)-1][m]/tmp);}}return 0;
}

Talented Chef

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3778

题意：给出n和m以及n个数，要求每次从n个数里取出m个数并减一，问最少需要几次把n个数减为0.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define eps 1e-8
using namespace std;int main()
{int test;scanf("%d",&test);while(test--){int n,m;int maxnum = -1,sum = 0;scanf("%d %d",&n,&m);for(int i = 1; i <= n; i++){int x;scanf("%d",&x);sum += x;maxnum = max(maxnum,x);}int ans = sum/m; if(sum%m != 0) ans++;if(ans < maxnum)printf("%d\n",maxnum);else printf("%d\n",ans);}return 0;
}

Paint the Grid Reloaded（dfs+bfs）

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3781

题意：给出一个n*m的矩阵，只有‘O’和‘X’，每次可以选择一个四连块进行翻转，问使矩阵全部变为‘O’或‘X’需要的最少步数。

思路：dfs找出每个四连块并进行标号，缩点建图，每个点到其他所有点都有一个最大距离，找到这样一个点它到其他所有点的最大距离最短，该值就是最少需要翻转的次数。

用邻接矩阵建图一直TLE，改成前向星就A了。。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;const int maxn = 1610;
const int maxm = 42;
const int INF = 0x3f3f3f3f;
int dir[4][2] = {{-1,0},{1,0},{0,1},{0,-1}};
int n,m;
char s[maxm];
int a[maxm][maxm],vis[maxm][maxn],scc;
int p[maxn],cnt;
int inque[maxn],dis[maxn];
queue <int> que;struct node
{int u,v,next;
}edge[100000];void init()
{memset(a,-1,sizeof(a));memset(vis,0,sizeof(vis));scc = 0;memset(p,-1,sizeof(p));cnt = 0;
}int judge(int i,int j)
{if(i >= 1 && i <= n && j >= 1 && j <= m)return 1;return 0;
}void add(int u, int v)
{edge[cnt] = (struct node){u,v,p[u]};p[u] = cnt++;
}void dfs(int i, int j, int col)
{if(vis[i][j] || a[i][j] != col)return;vis[i][j] = scc;dfs(i-1,j,col);dfs(i+1,j,col);dfs(i,j-1,col);dfs(i,j+1,col);
}void build()
{for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){for(int k = 0; k < 4; k++){int ii = i + dir[k][0];int jj = j + dir[k][1];if(judge(ii,jj) && vis[ii][jj] != vis[i][j])add(vis[i][j],vis[ii][jj]);}}}
}int bfs(int s)
{while(!que.empty()) que.pop();memset(inque,0,sizeof(inque));memset(dis,INF,sizeof(dis));int ans = 0;dis[s] = 0;inque[s] = 1;que.push(s);while(!que.empty()){int u = que.front();que.pop();for(int i = p[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(!inque[v]){inque[v] = 1;que.push(v);dis[v] = min(dis[v],dis[u]+1);ans = max(ans,dis[v]);}}}return ans;
}int main()
{int test;scanf("%d",&test);while(test--){init();scanf("%d %d",&n,&m);for(int i = 1; i <= n; i++){scanf("%s",s+1);for(int j = 1; j <= m; j++)if(s[j] == 'O')a[i][j] = 0;else a[i][j] = 1;}for(int i = 1; i <= n; i++){for(int j = 1; j <= m; j++){if(!vis[i][j]){scc++;dfs(i,j,a[i][j]);}}}build();int ans = INF;for(int i = 1; i <= scc; i++){int res = bfs(i);ans = min(ans,res);}printf("%d\n",ans);}return 0;
}