【poj-2195】 Going Home（二分图带权匹配+最小费用最大流）-优快云博客

本文链接：https://blog.youkuaiyun.com/x_Armin/article/details/99699770

描述

传送门：poj-2195

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.
Input

Input

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 $M$.

Output

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

Examples

intput

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

output
1
2
3
2
10
28

大致题意

m表示人，H表示房子，每个房子只能进一个人，房子数等于人数。要让所有人到自己的房子的距离和最小，请问这个距离和是多少。$(x_1,y_1)$和$(x_2,y_2)$之间的距离就是$|x_1 - x_2|+|y_1-y_2|$.

思路

加入超级源点s和超级汇点t，所有房子和s连一条边，所有人和t连一条边，从s到t跑最小费用最大流。

代码

#include<stdio.h>
#include<queue>
#include<string.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define repd(i,a,n) for(int i=n;i>=a;i--)
#define CRL(a,x) memset(a,x,sizeof(a))
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=1e3+5;
const int mod=1e9+7;
int abs(int x){return x<0? -x:x;}
int min(int a,int b){return a<b? a:b;}

int inq[N],dis[N],Begin[N],per[N],Flow[N],n,m,cnt;
struct node {int to,c,next,cost,f;}Edge[N*20];

void add(int u,int v,int c,int cost){
    Edge[++cnt]=(node){v,c,Begin[u],cost,0};
    Begin[u]=cnt;
    Edge[++cnt]=(node){u,0,Begin[v],-cost,0};
    Begin[v]=cnt;
}

bool spfa(int s,int t,int &flow,ll &cost){  //SPFA找一条s到t的费用最短路
    rep(i,s,t) dis[i]=inf;
    CRL(inq,0);dis[s]=0;
    per[s] = -1;
    Flow[s] = inf;

    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop(); inq[u] = 0;
        for(int i=Begin[u];~i;i=Edge[i].next){
            int v=Edge[i].to;
            if(Edge[i].c>Edge[i].f&&dis[v]>dis[u]+Edge[i].cost){
                dis[v]=dis[u]+Edge[i].cost;
                Flow[v]=min(Flow[u],Edge[i].c-Edge[i].f);
                per[v]=i;
                if(!inq[v]){
                    q.push(v);
                    inq[v]=1;
                }
            }
        }
    }
    if(dis[t]==inf) return false;
    flow += Flow[t];
    cost += (ll)dis[t]*(ll)Flow[t];     //费用=距离*流量
    for(int i=per[t];~i;i=per[Edge[i^1].to]){   //更新流量
        Edge[i].f+=Flow[t];
        Edge[i^1].f-=Flow[t];
    }
    return true;
}

int MCMF(int s,int t,ll &cost){
    cost =0;
    int flow = 0;
    while(spfa(s,t,flow,cost));
    return flow;
}

void Built(){       //建图
    int H[N][2],M[N][2],cntm=0,cnth=0,s=0,t=201;
    char x;
    rep(i,1,n){
        getchar();
        rep(j,1,m){
            x=getchar();
            if(x=='H') {
                H[++cnth][0]=i;
                H[cnth][1]=j;
            }
            else if(x=='m'){
                M[++cntm][0]=i;
                M[cntm][1]=j;
            }
        }
    }
    rep(i,1,cnth){
        add(s,i,1,0);
        add(i+100,t,1,0);
        rep(j,1,cntm)
            add(i,j+100,1,abs(H[i][0]-M[j][0])+abs(H[i][1]-M[j][1]));
    }
}

int main(){
    ll cost;int s=0,t=201;
    while(scanf("%d%d",&n,&m),n+m){
    	cnt=-1;
        CRL(Begin,-1);
        Built();
        MCMF(s,t,cost);
    	printf("%d\n",cost);
    }

    return 0;
}