[bzoj1433][ZJOI2009]假期的宿舍 二分图最大匹配

1433: [ZJOI2009]假期的宿舍

Time Limit: 10 Sec  Memory Limit: 162 MB
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Description

Input

Output

Sample Input

1
3
1 1 0
0 1 0
0 1 1
1 0 0
1 0 0

Sample Output

ˆ ˆ

HINT

对于30% 的数据满足1 ≤ n ≤ 12。
对于100% 的数据满足1 ≤ n ≤ 50,1 ≤ T ≤ 20。

Source

考这么水的题，愣是被我数组开小RE了8个点

#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>#define INF 0x7fffffffusing namespace std;const int N = 205;int last[N],S,T,n,h[N],q[N],cnt=1,cur[N],sign[N],ans,MT,inclass[N];struct Edge{int to,next,v;}e[N*40];void insert( int u, int v, int w ){e[++cnt].to = v; e[cnt].next = last[u]; e[cnt].v = w; last[u] = cnt;e[++cnt].to = u; e[cnt].next = last[v]; e[cnt].v = 0; last[v] = cnt;}bool bfs(){memset(h,-1,sizeof(h));int tail = 1, head = 0;q[0] = 0; h[0] = 0;while( tail != head ){int now = q[head++];for( int i = last[now]; i; i = e[i].next )if( h[e[i].to] == -1 && e[i].v ){h[e[i].to] = h[now] + 1;q[tail++] = e[i].to;}}return h[T] != -1;}int dfs( int x, int f ){int w,used=0;if( x == T ) return f;for( int i = cur[x]; i; i = e[i].next )if( h[e[i].to] == h[x] + 1 ){w = dfs(e[i].to,min(e[i].v,f-used));e[i].v -= w; e[i^1].v += w; used += w;if( e[i].v ) cur[x] = i; if( f == used ) return f;}if( !used ) h[x] = -1;return used;}void dinic(){while( bfs() ){for( int i = S; i <= T; i++ ) cur[i] = last[i];ans += dfs(S,INF);}}int main(){scanf("%d", &MT);while( MT-- ){scanf("%d", &n); S = 0; T = n*2+1; int yu = n; ans = 0;cnt = 1; memset(last,0,sizeof(last)); memset(sign,0,sizeof(sign));for( int i = 1; i <= n; i++ ) scanf("%d", &inclass[i]);for( int i = 1,x; i <= n; i++ ){ scanf("%d", &x); if( inclass[i] && x == 1 ) sign[i] = 1; }for( int i = 1; i <= n; i++ ) if( !sign[i] ) insert(S,i,1);for( int i = 1; i <= n; i++ ) if( inclass[i] ) insert(i+n,T,1);for( int i = 1; i <= n; i++ )for( int j = 1,x; j <= n; j++ ){scanf("%d", &x);if( sign[i] || !inclass[j] ) continue;if( x || i == j ) insert(i,j+n,1);}for( int i = 1; i <= n; i++ ) if( sign[i] ) yu--;dinic();if( ans == yu ) puts("^_^");else puts("T_T");}return 0;}