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

Submit Status Practice POJ 2195

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=;
const int maxm=;
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[];
struct node{
  int x,y;

}p[maxn],q[maxn];

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

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

        }
        flow+=Min;

      }

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

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

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

     }
     return ;
}

5634445	qwerqqq	D	Accepted	1188	110	G++	2650	5 min ago
5634440	qwerqqq	D	Wrong Answer			G++	2650	5 min ago
5634430	qwerqqq	D	Accepted	1188	125	G++	2650	6 min ago
5634424	qwerqqq	D	Accepted	1188	110	G++	2654	7 min ago
5634358	qwerqqq	D	Accepted	1188	125	G++	2654	12 min ago
5634345	qwerqqq	D	Accepted	1192	110	G++	2654	12 min ago
5634339	qwerqqq	D	Accepted	1188	110	G++	2655	12 min ago
5634334	qwerqqq	D	Accepted	1228	110	G++	2655	13 min ago
5634307	qwerqqq	D	Accepted	1540	110	G++	2656	16 min ago
5634289	qwerqqq	D	Wrong Answer			G++	2651	17 min ago
5634263	qwerqqq	D	Time Limit Exceeded			C++	2647	19 min ago
5634250	qwerqqq	D	Wrong Answer			G++	2647	21 min ago
5634209	qwerqqq	D	Wrong Answer			G++	3005	24 min ago
5634202	qwerqqq	D	Time Limit Exceeded			C++	3005	24 min ago
5631907	qwerqqq	D	Wrong Answer			G++	2662	14 hr ago