POJ2195(KB11-D 最小费用最大流)

Posted 2020-10-04 Penn000

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Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23515		Accepted: 11853

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

 

所有的房子和超级源点连边，容量为1，费用为0。

所有的人和超级汇点连边，容量为1，费用为0。

所有的房子和所有的人相互连边，容量为1，费用为房子和人的曼哈顿距离。

 

  1 //2017-08-24
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <queue>
  7 #include <cmath>
  8 
  9 using namespace std;
 10 
 11 const int N = 1000;
 12 const int M = 100000;
 13 const int INF = 0x3f3f3f3f;
 14 int head[N], tot;
 15 struct Edge{
 16     int to, next, c, w;//c为容量，w为单位费用
 17 }edge[M];
 18 
 19 void add_edge(int u, int v, int c, int w){
 20     edge[tot].c = c;
 21     edge[tot].w = w;
 22     edge[tot].to = v;
 23     edge[tot].next = head[u];
 24     head[u] = tot++;
 25 
 26     edge[tot].c = 0;
 27     edge[tot].w = -w;
 28     edge[tot].to = u;
 29     edge[tot].next = head[v];
 30     head[v] = tot++;
 31 }
 32 
 33 bool vis[N];
 34 int pre[N], dis[N];//pre记录路径，dis记录到源点的最小花费
 35 struct MinCostMaxFlow{
 36     int S, T;
 37     int flow, cost;
 38     void init(int _S, int _T){
 39         S = _S;
 40         T = _T;
 41         tot = 0;
 42         memset(head, -1, sizeof(head));
 43     }
 44     bool spfa(){
 45         memset(vis, 0, sizeof(vis));
 46         memset(dis, INF, sizeof(dis));
 47         dis[S] = 0;
 48         vis[S] = 1;
 49         queue<int> que;
 50         que.push(S);
 51         while(!que.empty()){
 52             int u = que.front();
 53             que.pop();
 54             for(int i = head[u]; i != -1; i = edge[i].next){
 55                 int v = edge[i].to;
 56                 if(edge[i].c > 0 && dis[v] > dis[u]+edge[i].w){
 57                     dis[v] = dis[u] + edge[i].w;
 58                     pre[v] = i;
 59                     if(!vis[v]){
 60                         vis[v] = true;
 61                         que.push(v);
 62                     }
 63                 }
 64             }
 65             vis[u] = 0;
 66         }
 67         return dis[T] != INF;
 68     }
 69     int dfs(int &flow){
 70         int u, p, sum = INF, ans = 0;
 71         for(u = T; u != S; u = edge[p^1].to){
 72             //记录路径上的最小流值
 73             p = pre[u];
 74             sum = min(sum, edge[p].c);
 75         }
 76         for(u = T; u != S; u = edge[p^1].to){
 77             p = pre[u];
 78             edge[p].c -= sum;
 79             edge[p^1].c += sum;
 80             ans += sum*edge[p].w;
 81         }
 82         flow += sum;
 83         return ans;
 84     }
 85     int maxflow(){
 86         cost = 0, flow = 0;
 87         while(spfa()){//寻找增广路并增广
 88             cost += dfs(flow);
 89         }
 90         return cost;
 91     }
 92 }mcmf;
 93 
 94 string grid[N];
 95 int x[N], y[N];
 96 
 97 int main()
 98 {
 99     std::ios::sync_with_stdio(false);
100     //freopen("inputD.txt", "r", stdin);
101     int n, m;
102     while(cin>>n>>m && (n || m)){
103         int cnt_m = 0, cnt_h = 0;
104         int s = N-2, t = N-3;
105         mcmf.init(s, t);
106         for(int i = 0; i < n; i++){
107             cin>>grid[i];
108             for(int j = 0; j < m; j++){
109                 if(grid[i][j] == ‘H‘){
110                     add_edge(s, cnt_h, 1, 0);
111                     x[cnt_h] = i;
112                     y[cnt_h++] = j;
113                 }
114             }
115         }
116         for(int i = 0; i < n; i++){
117             for(int j = 0; j < m; j++){
118                 if(grid[i][j] == ‘m‘){
119                     add_edge(cnt_h+cnt_m, t, 1, 0);
120                     for(int k = 0; k < cnt_h; k++){
121                         add_edge(k, cnt_h+cnt_m, 1, abs(i-x[k])+abs(j-y[k]));
122                     }
123                     cnt_m++;
124                 }
125             }
126         }
127         cout<<mcmf.maxflow()<<endl;
128     }
129 
130     return 0;
131 }