Going Home（最小费用最大流）

Going Home

http://poj.org/problem?id=2195

Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 26135   Accepted: 13106

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

 

 

最小费用最大流模板题

每个点连相邻的边建图   

  1 #include<iostream>
  2 #include<algorithm>
  3 #include<queue>
  4 #include<cstring>
  5 using namespace std;
  6 
  7 const int INF=0x3f3f3f3f;
  8 const int N=500005;
  9 const int M=500005;
 10 int top;
 11 int dist[N],pre[N];
 12 bool vis[N];
 13 int c[N];
 14 int maxflow;
 15 
 16 struct Vertex{
 17     int first;
 18 }V[N];
 19 struct Edge{
 20     int v,next;
 21     int cap,flow,cost;
 22 }E[M];
 23 
 24 void init(){
 25     memset(V,-1,sizeof(V));
 26     top=0;
 27     maxflow=0;
 28 }
 29 
 30 void add_edge(int u,int v,int c,int cost){
 31     E[top].v=v;
 32     E[top].cap=c;
 33     E[top].flow=0;
 34     E[top].cost=cost;
 35     E[top].next=V[u].first;
 36     V[u].first=top++;
 37 }
 38 
 39 void add(int u,int v,int c,int cost){
 40     add_edge(u,v,c,cost);
 41     add_edge(v,u,0,-cost);
 42 }
 43 
 44 bool SPFA(int s,int t,int n){
 45     int i,u,v;
 46     queue<int>qu;
 47     memset(vis,false,sizeof(vis));
 48     memset(c,0,sizeof(c));
 49     memset(pre,-1,sizeof(pre));
 50     for(i=1;i<=n;i++){
 51         dist[i]=INF;
 52     }
 53     vis[s]=true;
 54     c[s]++;
 55     dist[s]=0;
 56     qu.push(s);
 57     while(!qu.empty()){
 58         u=qu.front();
 59         qu.pop();
 60         vis[u]=false;
 61         for(i=V[u].first;~i;i=E[i].next){
 62             v=E[i].v;
 63             if(E[i].cap>E[i].flow&&dist[v]>dist[u]+E[i].cost){
 64                 dist[v]=dist[u]+E[i].cost;
 65                 pre[v]=i;
 66                 if(!vis[v]){
 67                     c[v]++;
 68                     qu.push(v);
 69                     vis[v]=true;
 70                     if(c[v]>n){
 71                         return false;
 72                     }
 73                 }
 74             }
 75         }
 76     }
 77     if(dist[t]==INF){
 78         return false;
 79     }
 80     return true;
 81 }
 82 
 83 int MCMF(int s,int t,int n){
 84     int d;
 85     int i,mincost;
 86     mincost=0;
 87     while(SPFA(s,t,n)){
 88         d=INF;
 89         for(i=pre[t];~i;i=pre[E[i^1].v]){
 90             d=min(d,E[i].cap-E[i].flow);
 91         }
 92         maxflow+=d;
 93         for(i=pre[t];~i;i=pre[E[i^1].v]){
 94             E[i].flow+=d;
 95             E[i^1].flow-=d;
 96         }
 97         mincost+=dist[t]*d;
 98     }
 99     return mincost;
100 }
101 string mp[105];
102 int dir[4][2]={0,1,1,0,0,-1,-1,0};
103 int main(){
104     int n,m;
105     int v,u,w,c;
106     int s,t;
107     while(cin>>n>>m){
108         if(!n&&!m) break;
109         init();
110         for(int i=0;i<n;i++){
111             cin>>mp[i];
112         }
113         s=0,t=n*m+1;
114         for(int i=0;i<n;i++){
115             for(int j=0;j<m;j++){
116                 if(mp[i][j]=='m'){
117                     add(s,i*m+j+1,1,0);
118                 }
119                 else if(mp[i][j]=='H'){
120                     add(i*m+j+1,t,1,0);
121                 }
122             }
123         }
124         int xx,yy;
125         for(int i=0;i<n;i++){
126             for(int j=0;j<m;j++){
127                 for(int k=0;k<4;k++){
128                     xx=i+dir[k][0];
129                     yy=j+dir[k][1];
130                     if(xx>=0&&xx<n&&yy>=0&&yy<m){
131                         add(i*m+j+1,xx*m+yy+1,INF,1);
132                     }
133                 }
134             }
135         }
136         int ans=MCMF(s,t,t+1);
137         cout<<ans<<endl;
138     }
139 }
View Code

 

posted on 2018-11-19 14:07  Fighting_sh  阅读(249)  评论(0)  编辑  收藏  举报

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