POJ 2195 Going Home 最小费用流

Going Home

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23975		Accepted: 12034

Description

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

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.

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

Source

Pacific Northwest 2004

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题意：给出一个图，m代表人，h代表房子，一间房子只能住一个人，格子之间距离都为1，问所有人走到房子最短距离是多少；

思路：可以把距离看成费用，这题就是最小费用流问题，房子代表汇点，人代表源点；

建图：点与点之间建边，费用为1，容量无限；

            超级源点s连向m，容量1，费用0；

            h连向超级汇点t，容量1，费用0；

            反向边费用相反；

AC code：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <queue>
#include <vector>
#include <stack>
#include <iostream>
#include <algorithm>

using namespace std;

const int MAXN = 20000+50;
const int MAXE = MAXN*4;
const int INF = 0x3f3f3f3f;

struct node {
    int u,v,w;
    int cap,next;
} edge[MAXE] ;

int pre[MAXN];
int Head[MAXN];
int Dis[MAXN];
bool vis[MAXN];
int n,m,top;

void Init(){
    memset(Head,-1,sizeof(Head));
    top=0;
}

void AddEdge(int u,int v,int w,int cap) {
    edge[top].u=u;  edge[top].v=v;
    edge[top].w=w;  edge[top].cap=cap;
    edge[top].next=Head[u];
    Head[u] = top++;

    edge[top].u=v;  edge[top].v=u;
    edge[top].w=-w; edge[top].cap=0;
    edge[top].next=Head[v];
    Head[v]=top++;
}

int SPFA(int s,int t) {
    stack<int> Q;
    memset(Dis,INF,sizeof Dis);
    memset(vis,false,sizeof vis);
    memset(pre,-1,sizeof pre);
    Dis[s] = 0 ; vis[s]=true;
    Q.push(s);
    while(!Q.empty()) {
        int u = Q.top(); Q.pop();
        for(int i = Head[u]; i!=-1; i = edge[i].next) {
            int v = edge[i].v;
            if(edge[i].cap && Dis[v]>Dis[u]+edge[i].w) {
                Dis[v] = Dis[u]+edge[i].w;
                pre[v]=i;
                if(!vis[v]) {
                    vis[v]=true;
                    Q.push(v);
                }
            }
        }
        vis[u] = false;
    }
    if(Dis[t]==INF) return 0;
    else return 1;
}

int MCF(int s,int t){
    int MinFlow,MinCost=0;
    while(SPFA(s,t)){

        MinFlow=INF;
        for(int k=pre[t];k!=-1;k=pre[edge[k].u]){
            MinFlow=min(MinFlow,edge[k].cap);
        }
        for(int k=pre[t];k!=-1;k=pre[edge[k].u]){
            edge[k].cap-=MinFlow;
            edge[k^1].cap+=MinFlow;
        }
        MinCost+=Dis[t]*MinFlow;
    }
    return MinCost;
}

int getIndex(int i,int j){
    return i*m+j;
}

int ok(int i,int j){
    if(i>=0&&i<n&&j>=0&&j<m) return 1;
    else return 0;
}

char map[200][200];
int dx[]={0,0,1,-1};
int dy[]={1,-1,0,0};

int main() {
    while(~scanf("%d%d",&n,&m)){
        if(n==0&&m==0) break;
        for(int i=0;i<n;i++){
            scanf("%s",&map[i]);
        }

        int s=n*m+1,t=n*m+2;
        Init();

        for(int i=0;i<n;i++){
            //printf("%s\n",map[i]);
            for(int j=0;j<m;j++){
                if(map[i][j]=='m') AddEdge(s,getIndex(i,j),0,1);
                else if(map[i][j]=='H') AddEdge(getIndex(i,j),t,0,1);
                for(int k=0;k<4;k++){
                    int x=i+dx[k];
                    int y=j+dy[k];
                    if(ok(x,y)) AddEdge(getIndex(i,j),getIndex(x,y),1,INF);
                }
            }
        }

        int ans=MCF(s,t);
        printf("%d\n",ans );
    }
    return 0;
}