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:\n%d\n",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*/