1059: [ZJOI2007]矩阵游戏
思路：
可以在对角线上填满的条件是，每一行，每一列，都要有一个数字。所以把列看成一排点，行看成一排点。二分图匹配，要求匹配数是n，即全匹配成功。
代码：
1 #include<cstdio> 2 #include<algorithm> 3 #include<cstring> 4 #include<cctype> 5 6 using namespace std; 7 8 const int N = 510; // 两倍空间！ 9 const int INF = 1e9; 10 struct Edge{ 11 int to,nxt,c; 12 Edge() {} 13 Edge(int x,int y,int z) {to = x;c = y;nxt = z;} 14 }e[100100]; 15 int head[N],dis[N],cur[N],q[100100]; 16 int S,T,tot,L,R; 17 18 inline int read() { 19 int x = 0,f = 1;char ch = getchar(); 20 for (; !isdigit(ch); ch=getchar()) if(ch=='-') f = -1; 21 for (; isdigit(ch); ch=getchar()) x = x*10+ch-'0'; 22 return x * f; 23 } 24 25 void init() { 26 tot = 1; 27 memset(head,0,sizeof(head)); 28 } 29 void add_edge(int u,int v,int w) { 30 e[++tot] = Edge(v,w,head[u]);head[u] = tot; 31 e[++tot] = Edge(u,0,head[v]);head[v] = tot; 32 } 33 bool bfs() { 34 for (int i=1; i<=T; ++i) cur[i] = head[i],dis[i] = -1; 35 L = 1;R = 0; 36 q[++R] = S; 37 dis[S] = 1; 38 while (L <= R) { 39 int u = q[L++]; 40 for (int i=head[u]; i; i=e[i].nxt) { 41 int v = e[i].to; 42 if (dis[v] == -1 && e[i].c > 0) { 43 dis[v] = dis[u] + 1; 44 q[++R] = v; 45 if (v==T) return true; 46 } 47 } 48 } 49 return false; 50 } 51 int dfs(int u,int flow) { 52 if (u == T) return flow; 53 int used = 0,tmp; 54 for (int &i=cur[u]; i; i=e[i].nxt) { 55 int v = e[i].to; 56 if (dis[v] == dis[u] + 1 && e[i].c > 0) { 57 tmp = dfs(v,min(e[i].c,flow-used)); 58 if (tmp) { 59 e[i].c -= tmp;e[i^1].c += tmp; 60 used += tmp; 61 if (used == flow) break; 62 } 63 } 64 } 65 if (used != flow) dis[u] = -1; 66 return used; 67 } 68 int dinic() { 69 int ans = 0; 70 while (bfs()) ans += dfs(S,INF); 71 return ans; 72 } 73 int main() { 74 int Case = read(); // 不要设成T 75 while (Case--) { 76 init(); 77 int n = read(); 78 S = n+n+1,T = n+n+2; 79 for (int i=1; i<=n; ++i) 80 for (int j=1; j<=n; ++j) { 81 int a = read(); 82 if (a) add_edge(i,j+n,1); 83 } 84 for (int i=1; i<=n; ++i) 85 add_edge(S,i,1),add_edge(i+n,T,1); 86 int ans = dinic(); 87 if (ans==n) puts("Yes"); 88 else puts("No"); 89 } 90 return 0; 91 }