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  • POJ 2195 Going Home 最小费用最大流 尼玛，心累

    D - Going Home
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
    Appoint description:
     

    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



    尼玛心累啊，为什么从源点链接房子，在链接人在链接会点jiu不行，，，，，，必须按照源点---》ren---》房子，，，，会点的顺序建图
    还有数组额外注意，
    尼玛了，因为数组开小，wa了14个小时，，，，，
    #include<stdio.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<vector>
#include<map>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn=10005;
const int maxm=20505;
const int inf=0x3f3f3f3f;
struct Edge{
   int to;
   int next;
   int cap,flow,cost;
}edge[maxm];
int head[maxn],tot;
int pre[maxn],dis[maxn];
bool vis[maxn];
int n;
char str[500];
struct node{
  int x,y;

}p[maxn],q[maxn];


void init(int nn){
    n=nn;
    tot=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost){
   edge[tot].to=v;
   edge[tot].cap=cap;
   edge[tot].cost=cost;
   edge[tot].flow=0;
   edge[tot].next=head[u];
   head[u]=tot++;
    edge[tot].to=u;
   edge[tot].cap=0;
   edge[tot].cost=-cost;
   edge[tot].flow=0;
   edge[tot].next=head[v];
   head[v]=tot++;
}
bool spfa(int s,int t){
      queue<int >q;
//    printf("n====%d
",n);
    for(int i=0;i<=n;i++){
         dis[i]=inf;
         vis[i]=false;
         pre[i]=-1;
    }
    dis[s]=0;
    vis[s]=true;
    q.push(s);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        vis[u]=false;
        for(int i=head[u];i!=-1;i=edge[i].next){
            int v=edge[i].to;
            if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost){
               dis[v]=dis[u]+edge[i].cost;
               pre[v]=i;
               if(!vis[v]){
                  vis[v]=true;
                  q.push(v);
               }

            }
        }


    }
    if(pre[t]==-1)
        return false;
    else
        return true;

}
int mincost(int s,int t,int &cost){
      int flow=0;
      cost=0;
    //  printf("%d
",head[2]);
      while(spfa(s,t)){
          int Min=inf;
        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){
            if(Min>edge[i].cap-edge[i].flow)
                Min=edge[i].cap-edge[i].flow;
        }
        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to]){
                edge[i].flow+=Min;
                edge[i^1].flow-=Min;
                cost+=edge[i].cost*Min;

        }
        flow+=Min;

      }

//      printf("cost=====%d  flow======%d
",cost,flow);
      return flow;
}

int main(){
     int x,y;
     while(scanf("%d%d",&x,&y)!=EOF){
             if(x==0&&y==0)
                 break;
             n=x*y+1;
             init(n);
             int start=0;
             int end=n;
             int pp=0,qq=0;
             for(int i=1;i<=x;i++){
                   scanf("%s",str+1);
                 for(int j=1;j<=y;j++){
                      int cnt=(i-1)*y+j;
                      if(str[j]=='m'){
                         addedge(start,cnt,1,0);
                         p[pp].x=i;
                         p[pp++].y=j;
                      }
                      if(str[j]=='H'){
                         addedge(cnt,end,1,0);
                         q[qq].x=i;
                         q[qq++].y=j;
                      }
                }
            //     printf("%s
",str+1);
             }

             for(int i=0;i<pp;i++){
                 for(int j=0;j<qq;j++){
                     addedge((p[i].x-1)*y+p[i].y,(q[j].x-1)*y+q[j].y,1,abs(p[i].x-q[j].x)+abs(p[i].y-q[j].y));
                 }
             }
         int tmp;
            int ans=mincost(start,end,tmp);
            printf("%d
",tmp);

     }
     return 0;
}
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  • 原文地址：https://www.cnblogs.com/13224ACMer/p/5268235.html
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