题目大意：一个农场中有狼和羊，现在要将他们用围栏分开，问最少需要多少围栏。

思路：所有源向所有狼连边，所有羊向汇连边，图中的每个相邻的格子之间连边，然后跑S->T的最大流，也就是把狼和羊分开的最小割。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 11000
#define MAXE 500010
#define INF 0x3f3f3f3f
#define S 0
#define T (m * n + 1)
using namespace std;
const int dx[] = {0,1,-1,0,0};
const int dy[] = {0,0,0,1,-1};

int m,n;
int src[110][110],num[110][110],cnt;

int head[MAXE],total = 1;
int next[MAXE],aim[MAXE],flow[MAXE];

int deep[MAXE];

inline void Add(int x,int y,int f)
{
	next[++total] = head[x];
	aim[total] = y;
	flow[total] = f;
	head[x] = total;
}

inline void Insert(int x,int y,int f)
{
	Add(x,y,f);
	Add(y,x,0);
}

inline bool BFS()
{
	static queue<int> q;
	while(!q.empty())	q.pop();
	memset(deep,0,sizeof(deep));
	deep[S] = 1;
	q.push(S);
	while(!q.empty()) {
		int x = q.front(); q.pop();
		for(int i = head[x]; i; i = next[i])
			if(!deep[aim[i]] && flow[i]) {
				deep[aim[i]] = deep[x] + 1;
				q.push(aim[i]);
				if(aim[i] == T)	return true;
			}
	}
	return false;
}

int Dinic(int x,int f)
{
	if(x == T)	return f;
	int temp = f;
	for(int i = head[x]; i; i = next[i])
		if(deep[aim[i]] == deep[x] + 1 && flow[i] && temp) {
			int away = Dinic(aim[i],min(flow[i],temp));
			if(!away)	deep[aim[i]] = 0;
			flow[i] -= away;
			flow[i^1] += away;
			temp -= away;
		}
	return f - temp;
}

int main()
{
	cin >> m >> n;
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j) {
			scanf("%d",&src[i][j]),num[i][j] = ++cnt;
			if(src[i][j] == 1)
				Insert(S,num[i][j],INF);
			if(src[i][j] == 2)
				Insert(num[i][j],T,INF);
		}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j)
			for(int k = 1; k <= 4; ++k) {
				int fx = i + dx[k];
				int fy = j + dy[k];
				if(!num[fx][fy])	continue;
				Insert(num[i][j],num[fx][fy],1);
			}
	int max_flow = 0;
	while(BFS())
		max_flow += Dinic(S,INF);
	cout << max_flow << endl;
	return 0;
}

BZOJ 1412 ZJOI 2009 狼和羊的故事 最小割