POJ2195:Going Home

題目連結：http://poj.org/problem?id=2195

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

題意：

給出一個n*m的棋盤，有些點上有房子，有些點上有人，最後人都要進房子裡（房子的數量和人的數量想等），每個人每走一步就有1的金幣。

問當所有人都進房子裡面時，所給出的金幣數量最少為多少。

題解：

求出每個人到房子的曼哈頓距離並建圖後，直接跑最大流最小費用流就好了。

程式碼如下：

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

const int N = 105 ,INF = 99999999,t = 505;
int n,m,cnt,tot,num;
int map[N][N],d[N][N],head[N*N],dis[N*N],vis[N*N],a[N*N],p[N*N],pre[N*N];
int Dis(int x1,int y1,int x2,int y2){
    return abs(x1-x2)+abs(y1-y2);
}
struct Edge{
    int u,v,next,c,w;
}e[N*N*N];
void adde(int u,int v,int w){
    e[tot].v=v;e[tot].c=1;e[tot].w=w;e[tot].next=head[u];head[u]=tot++;
    e[tot].v=u;e[tot].c=0;e[tot].w=-w;e[tot].next=head[v];head[v]=tot++;
}
int BellmanFord(int s,int t,int &flow,int &cost){
    for(int i=0;i<=t+5;i++) dis[i]=INF,a[i]=INF;
    queue <int> q;memset(vis,0,sizeof(vis));memset(p,-1,sizeof(p));
    memset(pre,-1,sizeof(pre));
    q.push(s);vis[s]=1;dis[s]=0;p[s]=0;
    while(!q.empty()){
        int u=q.front();q.pop();vis[u]=0;
        for(int i=head[u];i!=-1;i=e[i].next){
            int v=e[i].v;
            if(e[i].c>0 &&dis[v]>dis[u]+e[i].w){
                dis[v]=dis[u]+e[i].w;
                a[v]=min(a[u],e[i].c);
                p[v]=u;
                pre[v]=i;
                if(!vis[v]){
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
    if(dis[t]==INF) return 0;
    flow+=a[t];
    cost+=a[t]*dis[t];
    for(int i=t;i>0;i=p[i]){//
        e[pre[i]].c-=a[t];
        e[pre[i]^1].c+=a[t];
    }
    return 1;
}
int Min_cost(){
    int flow=0,cost=0;
    while(BellmanFord(0,t,flow,cost));
    return cost;
}
int main(){
    while(~scanf("%d%d",&n,&m)){
        if(n==0 && m==0) break;
        tot=0;num=0;cnt=0;
        memset(head,-1,sizeof(head));
        vector <pair<int,int> > h;
        memset(map,0,sizeof(map));
        for(int i=1;i<=n;i++){
            char s[N];
            scanf("%s",s);
            for(int j=0;j<m;j++){
                if(s[j]=='H'){
                    map[i][j+1]=2;
                    h.push_back(make_pair(i,j+1));
                }
                if(s[j]=='m') map[i][j+1]=1,num++;
            }
        }
        for(int i=1;i<=num;i++) adde(0,i,0);
        for(int i=num+1;i<=2*num;i++) adde(i,t,0);
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                if(map[i][j]!=1) continue ;
                cnt++;int tmp=0;
                for(int k=0;k<h.size();k++){
                    tmp++;
                    pair <int,int> v = h[k];
                    adde(cnt,num+tmp,Dis(i,j,v.first,v.second));
                }
            }
        }
        printf("%d\n",Min_cost());
    }
    return 0;
}