BZOJ1433 假期的宿舍(最大流)

是一張二分圖，也可以用匈牙利算法實現。

X部爲人，Y部爲牀，S向要留校的和校外人員連容量爲1的邊，所有的牀向T連邊，只有校內人員有牀。

#include<cstdio>
#include<cstring>
#include<vector>
#define MAXN 110
#define INF 0x3f3f3f3f
using namespace std;
inline int Min(int a,int b)
{return a<b?a:b;}
struct E
{
	int v,w,op;
	E(){}
	E(int a,int b,int c)
	{v = a; w = b; op = c;}
};
vector<E> g[MAXN];
void add(int u,int v,int w)
{
	g[u].push_back(E(v,w,g[v].size()));
	g[v].push_back(E(u,0,g[u].size()-1));
}
int d[MAXN],vd[MAXN],n,s,t;
int sc[MAXN],lv[MAXN],a,sum,flow;
int aug(int i,int augco)
{
	int j,mind = t-1,delta,augc = augco,sz = g[i].size();
	if(i == t) return augco;

	for(j = 0; j < sz; j++)
	{
		int v = g[i][j].v;
		if(g[i][j].w)
		{
			if(d[i] == d[v]+1)
			{
				delta = Min(augc,g[i][j].w);
				delta = aug(v,delta);
				g[i][j].w -= delta;
				g[v][g[i][j].op].w += delta;
				augc -= delta;
				if(d[s] >= t) return augco - augc;
				if(!augc) break;
			}
			if(d[v] < mind) mind = d[v];
		}
	}
	if(augco == augc)
	{
		vd[d[i]]--;
		if(!vd[d[i]]) d[s] = t;
		d[i] = mind+1;
		vd[d[i]]++;
	}
	return augco - augc;
}
void sap()
{
	flow = 0;
	memset(d,0,sizeof d);
	memset(vd,0,sizeof vd);
	vd[0] = t;
	while(d[s] < t)
		flow += aug(s,0x3f3f3f3f);
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		sum = 0;
		memset(sc,0,sizeof sc);
		memset(lv,0,sizeof lv);
		scanf("%d",&n);
		s = 2*n+1,t = s+1;
		for(int i = 1; i <= n; i++) 
		{	
			scanf("%d",&sc[i]);
			if(sc[i]) add(i+n,t,1);
		}	
		for(int i = 1; i <= n; i++)
		{
			scanf("%d",&lv[i]);
			if((sc[i]&&!lv[i])||!sc[i]) add(s,i,1),sum++;
		}
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
			{
				scanf("%d",&a);
				if(a||i == j) add(i,j+n,INF);
			}
		sap();
		if(flow == sum) printf("^_^\n");
		else printf("T_T\n");
		for(int i = 1; i <= t; i++) g[i].clear();
	}
}