Going Home 费用流

题目：

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man. 

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a '.' means an empty space, an 'H' represents a house on that point, and am 'm' indicates there is a little man on that point. 

 You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

输入：

There are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of 'H's and 'm's on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.

输出：

For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.

样例输入：

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

样例输出：

2
10
28

题意：

m代表人，H代表房子，每个人要进入一个房子，只能水平或者垂直移动，移动一步需要支付一美元，计算需要支付的最低金额，最小费用最大流。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=101000,INF=0x7fffffff;
int head[maxn],next[maxn<<1],d[maxn],dis[maxn],cnt;
int p[maxn],f[maxn];
int s,t,flow,value;
char ch[105][105];
struct edge
{
    int u,v,w,c;
}node[maxn<<1];
struct NODE
{
    int x,y;
}man[maxn],house[maxn];
void add(int u, int v, int c, int w)
{
    node[cnt].u=u;
    node[cnt].v=v;
    node[cnt].c=c;
    node[cnt].w=w;
    next[cnt]=head[u];
    head[u]=cnt++;
}
bool spfa()
{
    deque<int>q;
    for(int i=0;i<maxn;i++) d[i]=INF;

    memset(dis, 0 ,sizeof (dis));
    memset(p,-1,sizeof(p));
    q.push_front(s) ;
    d[s]=0;
    dis[s]=1;
    p[s]=0;
    f[s]=INF;
    while(!q.empty())
    {
        int u=q.front();q.pop_front();
        dis[u]=0;
        for(int i=head[u];i!=-1;i=next[i])
        {
            edge e=node[i];
            if(d[e.v]>d[e.u]+e.w && e.c > 0)
            {
                d[e.v] = d[e.u] + e.w;
                p[e.v] =i;
                f[e.v] = min(f[u], e.c);
                if(!dis[e.v])
                {
                   if(q.empty())q.push_front(e.v);
                   else
                   {
                       if(d[e.v]<d[q.front()])q.push_front(e.v);
                       else q.push_back(e.v);
                   }
                    dis[e.v]=1;//不要忘记标记
                }
            }
        }
    }
    if(p[t]==-1) return 0;
    flow+=f[t];value+=f[t]*d[t];
    for(int i=t;i!=s;i=node[p[i]].u)
    {
        node[p[i]].c-=f[t];
        node[p[i]^1].c+=f[t];
    }
    return 1;
}
int maxflow()
{
    value=0;
    flow=0;
    while(spfa());
    return value;
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m) && n+m)
    {
        getchar();
        memset(head,-1,sizeof(head));
        cnt=0;
        int ans=0;
        int summan=0;//人的个数    
        int sumhouse=0;//房子的个数
        for(int i=0;i<n;i++)
        {
            scanf("%s",ch[i]);
            for(int j=0;j<m;j++)
            {

                if(ch[i][j] == 'm')
                {
                    summan++;
                    man[summan].x = i;//给每个人存下坐标，方便后面求距离
                    man[summan].y = j;
                }
                if(ch[i][j]=='H')
                {
                    sumhouse++;
                    house[sumhouse].x=i;//给每个房子存下坐标，作用同上
                    house[sumhouse].y=j;
                }
            }
        }
        s=0;//超级源点和汇点
        t=summan+sumhouse+1;
        for(int i=1;i<=summan;i++)
        {
            add(s,i,1,0);//源点和人连接。容量为1，费用为0
            add(i,s,0,0);
        }
        for(int i=1;i<=summan;i++)
            for(int j=1;j<=sumhouse;j++)
        {
            int distant=abs(house[j].x-man[i].x)+abs(house[j].y-man[i].y);
            add(i,summan+j,1,distant);//人和房子连接，容量为1，费用为距离
            add(summan+j,i,0,-distant);
        }
        for(int i=1;i<=sumhouse;i++)
        {
            add(summan+i,t,1,0);//房子和汇点连接，容量为1，费用为0
            add(t,summan+i,0,0);
        }
        ans+=maxflow();
        printf("%d\n",ans);
    }
}