POJ 2195 Going Home（网络流-费用流）

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

Output

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

题目大意：

m表示人，H表示房子，它们之间的距离是曼哈顿距离，问你所有人一人个房子的总花费是多少？

用最小费用流即可。构图略。

#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int maxn=11000;
const int maxm=1100000;
const int inf=(1<<30);

struct edge{
    int u,v,next,f,c;
    edge(int u0=0,int v0=0,int f0=0,int c0=0,int next0=0){
        u=u0,v=v0,f=f0,c=c0,next=next0;
    }
}e[maxm];

int head[maxn],path[maxn],dist[maxn];
bool visited[maxn];
int cnt,src,sink;

void init(){
    cnt=0;
    memset(head,-1,sizeof(head));
}

void adde(int u,int v,int f,int c){
    e[cnt].u=u,e[cnt].v=v,e[cnt].f=f,e[cnt].c=c,e[cnt].next=head[u],head[u]=cnt++;
    e[cnt].u=v,e[cnt].v=u,e[cnt].f=0,e[cnt].c=-c,e[cnt].next=head[v],head[v]=cnt++;
}

bool bfs(){
     for(int i=src;i<=sink;i++){
        dist[i]=inf;
        path[i]=-1;
     }
     dist[src]=0;
     queue <int> q;
     q.push(src);
     visited[src]=true;
     while(!q.empty()){
         int s=q.front();
         q.pop();
         for(int i=head[s];i!=-1;i=e[i].next){
              int d=e[i].v;
              if(e[i].f>0 && dist[s]+e[i].c<dist[d]){
                 dist[d]=dist[s]+e[i].c;
                 path[d]=i;
                 if(!visited[d]){
                    visited[d]=true;
                    q.push(d);
                 }
              }
         }
         visited[s]=false;
     }
     return path[sink]>=0;
}

int getMinCost(){
    int ret=0;
    while(bfs()){
        int delta=inf;
        for(int i=sink;i!=src;i=e[path[i]].u){
            if( e[path[i]].f<delta ) delta=e[path[i]].f;
        }
        for(int i=sink;i!=src;i=e[path[i]].u){
            e[path[i]].f-=delta;
            e[path[i]^1].f+=delta;
        }
        ret+=dist[sink]*delta;
    }
    return ret;
}

int n,m;

void input(){
    init();
    src=0;
    char a[110];
    vector < pair<int,int> > house,man;
    for(int i=0;i<n;i++){
        scanf("%s",&a);
        for(int j=0;j<m;j++){
            if(a[j]=='H') house.push_back(make_pair(i,j));
            else if(a[j]=='m') man.push_back(make_pair(i,j));
        }
    }
    sink=house.size()+man.size()+1;
    for(int i=1;i<=man.size();i++) adde(src,i,1,0);
    for(int i=1;i<=man.size();i++){
        for(int j=1;j<=house.size();j++){
            int tdis=abs(man[i-1].first-house[j-1].first)+abs(man[i-1].second-house[j-1].second);
            adde(i,j+man.size(),1,tdis);
        }
    }
    for(int i=1;i<=house.size();i++) adde(i+man.size(),sink,1,0);
}

void solve(){
    printf("%d\n",getMinCost());
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF && (m||n) ){
        input();
        solve();
    }
    return 0;
}

POJ 2195 Going Home（网络流-费用流）,布布扣,bubuko.com

POJ 2195 Going Home（网络流-费用流）

原文：http://blog.csdn.net/a1061747415/article/details/38402111