POJ 2195 一人一房 最小费用流 建图 水题

Going Home

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

题意：

比较好理解，给定一个N*M的地图，地图上有若干个man和house，且man与house的数量一致。man每移动一格需花费$1（即单位费用=单位距离），一间house只能入住一个man。现在要求所有的man都入住house，求最小费用。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a,b) memset(a,b,sizeof(a))
typedef long long ll;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
const int big=50000;
int max(int a,int b) {return a>b?a:b;};
int min(int a,int b) {return a<b?a:b;};
const int N = 70;
const int M=10000+100;
struct edge{
   int to,cap,cost,rev;
};

vector<edge> G[1005];
int dist[1005],inq[1005],prev[1005],prel[1005];
struct Point{
   int x,y;
};

int n,m,x,y,c,cnth,cntm;
Point house[105],men[105];

void add_edge(int u,int v,int cap,int cost)
{
    G[u].push_back(edge{v,cap,cost,G[v].size()});
    G[v].push_back(edge{u,0,-cost,G[u].size()-1});
}

int mincost(int s,int t,int f)
{
    int ans=0;
    while(f>0)
    {
      memset(dist,inf,sizeof(dist));
      memset(inq,0,sizeof(inq));
      dist[s]=0;
      queue<int> q;
      q.push(s);
      inq[s]=1;
      MM(prev,-1);
      while(!q.empty())
      {
          int u=q.front();
          q.pop();inq[u]=0;
          for(int j=0;j<G[u].size();j++)
            {
                edge &e=G[u][j];
                if(e.cap>0&&dist[e.to]>dist[u]+e.cost)
                   {
                       dist[e.to]=dist[u]+e.cost;
                       prev[e.to]=u;
                       prel[e.to]=j;
                       if(!inq[e.to])
                       {
                           q.push(e.to);
                           inq[e.to]=1;
                       }
                   }
            }
       }
       for(int i=t;i>s;)
       {
           int f=prev[i];
           if(f==-1) return -1;
           int j=prel[i];
           G[f][j].cap-=1;
           G[i][G[f][j].rev].cap+=1;
           ans+=G[f][j].cost;
           i=prev[i];
       }
       f-=1;
    }
    return ans;
}

void build()
{
    for(int i=0;i<=cnth+cntm+1;i++) G[i].clear();
    for(int i=1;i<=cntm;i++)
          add_edge(0,i,1,0);
    for(int i=1;i<=cnth;i++)
          add_edge(cntm+i,cntm+cnth+1,1,0);
    for(int i=1;i<=cntm;i++)
          for(int j=1;j<=cnth;j++)
        {
          int dis=abs(house[j].x-men[i].x)+abs(house[j].y-men[i].y);
          add_edge(i,j+cntm,1,dis);
        }
}

char s[105][105];
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m)&&(n||m))
    {
        cnth=cntm=0;
        for(int i=1;i<=n;i++)
           {
               scanf("%s",s[i]);
               for(int j=0;j<m;j++)
                   if(s[i][j]==‘H‘)
                      house[++cnth]=(Point){i,j+1};
                   else if(s[i][j]==‘m‘)
                      men[++cntm]=(Point){i,j+1};
           }
        build();
        printf("%d\n",mincost(0,cnth+cntm+1,cnth));
    }
    return 0;
}

　　分析：建图有一点点技术含量，人放左边，门放右边，加一个源点与汇点，然后跑一跑大小为人数（门数）的最小费用流就好，