Going Home HDU

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.

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

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

Sample Output

2
10
28

题意：使得每个m都对应一个H时的最短的路径

题解：找一个源点，连接所有的m，再找一个汇点，使得所有H指向这个汇点。

然后跑一边最大费用最小流就ok了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <cmath>
using namespace std;
typedef long long ll;
const int MAXX=60010;
const int INF=0x3f3f3f3f;

struct node
{
    int st;
    int to;
    int next;
    int cap;
    int cost;
}edge[MAXX];

char mp[102][102];
int head[MAXX],tol;
int pre[MAXX],dis[MAXX];
bool vis[MAXX];
int n,m,p;

struct node1
{
    int x,y;
    node1(){}
    node1(int a,int b)
    {
        x=a;
        y=b;
    }
};
vector<node1> v1,v2;

void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
    v1.clear();
    v2.clear();
}

void addedge(int u,int v,int cap,int cost)
{
    edge[tol].st=u;
    edge[tol].to=v;
    edge[tol].cap=cap;
    edge[tol].cost=cost;
    edge[tol].next=head[u];
    head[u]=tol++;

    edge[tol].st=v;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].cost=-cost;
    edge[tol].next=head[v];
    head[v]=tol++;
}

bool SPFA(int s,int t)
{
    queue<int> q;
    memset(dis,INF,sizeof(dis));
    memset(vis,0,sizeof(vis));
    memset(pre,-1,sizeof(pre));
    dis[s]=0;
    vis[s]=1;
    q.push(s);
    while(!q.empty())
    { 
        int u=q.front(); q.pop();
        vis[u]=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int to=edge[i].to; 
            if(edge[i].cap>0&&dis[to]>dis[u]+edge[i].cost)
            {
                dis[to]=dis[u]+edge[i].cost;
                pre[to]=i;
                if(!vis[to])
                {
                    vis[to]=1;
                    q.push(to);
                }
            }
        }
    }
    if(pre[t]==-1)return 0;
    return 1;
}

int minCostMaxFlow(int s,int t)
{
    int cost=0;
    while(SPFA(s,t))
    { 
        int minn=INF;
        for(int i=pre[t];i!=-1;i=pre[edge[i].st])
            minn=min(minn,edge[i].cap);

        for(int i=pre[t];i!=-1;i=pre[edge[i].st])
        {
            edge[i].cap-=minn;
            edge[i^1].cap+=minn;
        }
        cost+=minn*dis[t];
    }
    return cost;
}

int main()
{
    while(scanf("%d%d",&n,&m)&&m&&n)
    {
        getchar();
        init();
        for(int i=0;i<n;i++)
        {
            scanf("%s",mp[i]);
            getchar();
        }
        for(int i=0;i<n;i++)
        for(int j=0;j<m;j++)
        {
            if(mp[i][j]=='m')
                v1.push_back(node1(i,j));
            if(mp[i][j]=='H')
                v2.push_back(node1(i,j));

        }
        int l=v1.size();
        int r=v2.size();
        for(int i=0;i<v1.size();i++)
        {

            node1 N1=v1[i];
            int x=i+1;
            addedge(0,x,1,0);
            for(int j=0;j<v2.size();j++)
            {
                node1 N2=v2[j];
                int y=j+l+1;
                int D=abs(N1.x-N2.x)+abs(N1.y-N2.y); 
                addedge(x,y,1,D);
                addedge(y,x,1,D);
                if(i==l-1)
                addedge(y,l+r+1,1,0);
            }
        }
        int ans=minCostMaxFlow(0,l+1+r);
        printf("%d
",ans);
    }
    return 0;
}