HDU 3046 Pleasant sheep and big big wolf 最小割

题意：

给定n*m个点的矩阵

0为空点。1为羊。2为狼

相邻点之间有一条路。

问要使得狼与羊不连通最少要去掉几条边

最小割

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;

#define ll int 

#define N 40010
#define M 105000
#define inf 1073741824
#define inf64 1152921504606846976
struct Edge{  
	ll from, to, cap, nex, max;  
}edge[M*4];//注意这个一定要够大 不然会re 还有反向弧  

ll head[N], edgenum;  
void add(ll u, ll v, ll cap){  
	Edge E = { u, v, cap, head[u],cap};  
	edge[ edgenum ] = E;  
	head[u] = edgenum ++;  

	Edge E2= { v, u, 0,   head[v],cap};  
	edge[ edgenum ] = E2;  
	head[v] = edgenum ++;  
}  
ll sign[N];  
bool BFS(ll from, ll to){  
	memset(sign, -1, sizeof(sign));  
	sign[from] = 0;  

	queue<ll>q;  
	q.push(from);  
	while( !q.empty() ){  
		int u = q.front(); q.pop();  
		for(ll i = head[u]; i!=-1; i = edge[i].nex)  
		{  
			ll v = edge[i].to;  
			if(sign[v]==-1 && edge[i].cap)  
			{  
				sign[v] = sign[u] + 1, q.push(v);  
				if(sign[to] != -1)return true;  
			}  
		}  
	}  
	return false;  
}  
ll Stack[N], top, cur[N];  
ll dinic(ll from, ll to){  

	ll ans = 0;  
	while( BFS(from, to) )  
	{  
		memcpy(cur, head, sizeof(head));  
		ll u = from;      top = 0;  
		while(1)  
		{  
			if(u == to)  
			{  
				ll flow = inf, loc;//loc 表示 Stack 中 cap 最小的边  
				for(ll i = 0; i < top; i++)  
					if(flow > edge[ Stack[i] ].cap)  
					{  
						flow = edge[Stack[i]].cap;  
						loc = i;  
					}  

					for(ll i = 0; i < top; i++)  
					{  
						edge[ Stack[i] ].cap -= flow;  
						edge[Stack[i]^1].cap += flow;  
					}  
					ans += flow;  
					top = loc;  
					u = edge[Stack[top]].from;  
			}  
			for(ll i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标  
				if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;  
			if(cur[u] != -1)  
			{  
				Stack[top++] = cur[u];  
				u = edge[ cur[u] ].to;  
			}  
			else  
			{  
				if( top == 0 )break;  
				sign[u] = -1;  
				u = edge[ Stack[--top] ].from;  
			}  
		}  
	}  
	return ans;  
}  

void init(){memset(head,-1,sizeof head);edgenum = 0;}

int mp[205][205];
int n, m;
int Hash(int x,int y){return (x-1)*m+y;}
bool inmap(int x,int y){return 1<=x&&x<=n&&1<=y&&y<=m;}
int w[N], s[N], step[4][2] = {1,0,-1,0,0,1,0,-1};
int main()
{
    int u, v, d, i, j, Cas = 1;
	while(~scanf("%d %d",&n,&m)){
		init();
		int wt = 0, st = 0;
		for(i = 1; i <= n; i++)
		{
			for(j = 1; j <= m; j++)
			{
				scanf("%d",&mp[i][j]);
				int now = Hash(i,j);
				if(mp[i][j]==1)s[st++] = now;
				if(mp[i][j]==2)w[wt++] = now;
				for(int k = 0; k < 4; k++){
					if(!inmap(i+step[k][0], j+step[k][1]))continue;
					int go = Hash(i+step[k][0], j+step[k][1]);
					add(now, go, 1);
				}
			}
		}
		int from = 0, to = Hash(n,m)+1;
		for(i = 0; i < wt; i++)add(from, w[i], inf);
		for(i = 0; i < st; i++)add(s[i], to, inf);
		printf("Case %d:
%d
",Cas++,dinic(from, to));
	}
    return 0;
}
/*
4 6
1 0 0 1 0 0
0 1 1 0 0 0
2 0 0 0 0 0
0 2 0 1 1 0

*/