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题目：在一个N * M 的矩阵草原上，分布着羊和狼，每个格子只能存在0或1只动物。现在要用栅栏将所有的狼和羊分开，问怎么放，栅栏数放的最少，求出个数？

解析：将狼群看作一个集合，羊群看作一个集合。然后设置源点和汇点，将两点至存在动物的点的距离赋值为1，构图，由于求得是栅栏数，从存在动物的位置向四周发散点赋值为1，即该方向放置一个栅栏。然后可以发现变成了求最小割，即求出最大流。需要注意的是，由于数据比较大，200 * 200。如果设置源点和汇点相差较大（即s = 0,e = n * m )，容易TLE.

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
 
// 最大流  ISAP + bfs+栈优化

const int maxn = 100010; //点的个数
const int maxm = 400010; //边的个数
const int INF = 0xfffffff;

struct Edge{
	int to, next, cap, flow;
}edge[ maxm ];//注意是maxm

int tol;
int head[ maxn ];
int gap[ maxn ], dep[ maxn ], cur[ maxn ];

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

void addedge( int u, int v, int w, int rw = 0 ){
	edge[ tol ].to = v; edge[ tol ].cap = w; edge[ tol ].flow = 0;
	edge[ tol ].next = head[ u ]; head[ u ] = tol++;
	edge[ tol ].to = u; edge[ tol ].cap = rw; edge[ tol ].flow = 0;
	edge[ tol ].next = head[ v ]; head[ v ] = tol++;
}
	 
int Q[ maxn ];

void BFS( int start, int end ){
	memset( dep, -1, sizeof( dep ) );
	memset( gap, 0, sizeof( gap ) );
	gap[ 0 ] = 1;
	int front = 0, rear = 0;
	dep[ end ] = 0;
	Q[ rear++ ] = end;
	while( front != rear ){
		int u = Q[ front++ ];
		for( int i = head[ u ]; i != -1; i = edge[ i ].next ){
			int v = edge[ i ].to;
			if( dep[ v ] != -1 )
				continue;
			Q[ rear++ ] = v;
			dep[ v ] = dep[ u ] + 1;
			gap[ dep[ v ] ]++;
		}
	}
}

int S[ maxn ];
int sap( int start, int end, int N ){
	BFS( start, end );
	memcpy( cur, head, sizeof( head ) );
	int top = 0;
	int u = start;
	int ans = 0;
	while( dep[ start ] < N ){
		if( u == end ){
			int Min = INF;
			int inser;
			for( int i = 0; i < top; ++i ){
				if( Min > edge[ S[ i ] ].cap - edge[ S[ i ] ].flow ){
					Min = edge[ S[ i ] ].cap - edge[ S[ i ] ].flow;
					inser = i;
				}
			}
			for( int i = 0; i < top; ++i ){
				edge[ S[ i ] ].flow += Min;
				edge[ S[ i ] ^ 1 ].flow -= Min;
			}
			ans += Min;
			top = inser;
			u = edge[ S[ top ] ^ 1 ].to;
			continue;
		}
		bool flag = false;
		int v;
		for( int i = cur[ u ]; i != -1; i = edge[ i ].next ){
			v = edge[ i ].to;
			if( edge[ i ].cap - edge[ i ].flow && dep[ v ] + 1 == dep[ u ] ){
				flag = true;
				cur[ u ] = i;
				break;
			}
		}
		if( flag ){
			S[ top++ ] = cur[ u ];
			u = v;
			continue;
		}
		int Min = N;
		for( int i = head[ u ]; i != -1; i = edge[ i ].next ){
			if( edge[ i ].cap - edge[ i ].flow && dep[ edge[ i ].to ] < Min ){
				Min = dep[ edge[ i ].to ];
				cur[ u ] = i;
			}
		}
		gap[ dep[ u ] ]--;
		if( !gap[ dep[ u ] ] )
			return ans;
		dep[ u ] = Min + 1;
		gap[ dep[ u ] ]++;
		if( u != start )
			u = edge[ S[ --top ] ^ 1 ].to;
	}
	return ans;
}
	 
int dir[ ][ 2 ] = { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } };

int main(){
	int N, M, Case = 1;
	while( scanf( "%d%d", &N, &M ) != EOF ){
		int s = N * M, t = N * M + 1;
		int n = N * M + 2, temp;
		init();
		for( int i = 0; i < N; ++i ){
			for( int j = 0; j < M; ++j ){
				scanf( "%d", &temp );
				if( temp == 2 ){
					addedge( s, M * i + j, INF );
				}
				if( temp == 1 ){
					addedge( M * i + j, t, INF );
				}
				for( int k = 0; k < 4; ++k ){
					int x = i + dir[ k ][ 0 ];
					int y = j + dir[ k ][ 1 ];
					if( x >= 0 && x < N && y >= 0 && y < M )
						addedge( M * i + j, M * x + y,  1 );
				}
			}
		}
		printf( "Case %d:\n%d\n", Case++, sap( s, t, n ) );
	}
	return 0;
}