POJ 2195

题目链接：Going Home

二分图最大权匹配

 KM  或者 最小费用最大流 

现在提供网络流的做法 

阅读全文请移步 费用流解最大权匹配

这篇博客也不错 二分图带权匹配

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
using namespace std;
#define ABSs(x) ((x)<0?-(x):(x))
const int N = 105;
struct Node{
	int ne;
	int to;
	int w;
	int co;
}e[N*(N+2)<<1+5];
int head[N<<1+6],hh[N*N],mm[N*N];
int path[N<<1+5],v[N<<1+5],dis[N<<1+5];
int n,m,len,st,en,cnt;
char s[N];
void init()
{
	memset(head,-1,sizeof(head));
	cnt =  0;
}
void add(int u,int v,int co,int val)
{
	e[cnt].to = v;
	e[cnt].w = val;
	e[cnt].co = co;
	e[cnt].ne = head[u];
	head[u] = cnt ++;
}
bool spfa(int st,int en)
{
	memset(path,-1,sizeof(path));
	memset(v,0,sizeof(v));
	memset(dis,127,sizeof(dis));
	queue<int>q;
	q.push(st);
	v[st] = 1;
	dis[st] = 0;
	while(!q.empty())
	{
		int now = q.front();
		q.pop();
		for(int i=head[now];~i;i=e[i].ne)
		{
			int to = e[i].to;
			if(e[i].w > 0 && dis[to] > dis[now] + e[i].co)
			{
				dis[to] = dis[now] + e[i].co;
				path[to] = i;
				if(!v[to])
				{
					v[to] = 1;
					q.push(to);
				}
			}
		}
		v[now] = 0;
	}
	return dis[en] != dis[N<<1+4];
}
int MCMF(int st,int en)
{
	int ans = 0;
	while(spfa(st,en))
	{
		int minn = 1e9;
		for(int i=path[en];~i;i=path[e[i^1].to])
		{
			if(minn > e[i].w)
			minn = e[i].w;
		}
		ans += minn*dis[en];
		for(int i=path[en];~i;i=path[e[i^1].to])
		{
			e[i].w -= minn;
			e[i^1].w += minn;
		}
	}
	return ans;
}
int main()
{
	while(scanf("%d%d",&n,&m) && n+m)
	{
		init();
		int hlen = 0;
		int mlen = 0;
		for(int i=0;i<n;i++)
		{
			scanf("%s",s);
			for(int j=0;j<m;j++)
			{
				if(s[j] == 'H')
				{
					hh[hlen++] = i*m+j;
				}
				else if(s[j] == 'm')
				{
					mm[mlen ++] = i*m+j;
				}
			}
		}
		int x0,x1,y0,y1;
		for(int i=0;i<mlen;i++)
		{
			for(int j=0;j<hlen;j++)
			{
				x0 = mm[j]/m;
				y0 = mm[j]%m;
				x1 = hh[i]/m;
				y1 = hh[i]%m;
				add(i+1,j+mlen+1,ABSs(x0-x1) + ABSs(y0-y1),1);
				add(j+mlen+1,i+1,- ABSs(x0-x1) - ABSs(y0-y1),0);
			}
		}
		n = mlen;
		m = hlen;
		st = 0;
		en = n + m + 1;
		for(int i=0;i<n;i++)
		{
			add(st,i+1,0,1);
			add(i+1,st,0,0);
		}
		for(int j=0;j<m;j++)
		{
			add(j+n+1,en,0,1);
			add(en,j+n+1,0,0);
		}
		printf("%d\n",MCMF(st,en));
	}
	return 0;
}