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hdu2195Going Home【最小费用最大流】

2024-07-06 05:20:36   232人阅读

Going Home
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 17230   Accepted: 8781

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：
5 5
HH..m
.....
.....
.....
mm..H
对于一个图
m代表人H代表房间
现在需要用最少的步数将所有的人送回房间，问最小步数是多少，每个房间只能容纳一个人


分析：
每个人到每个房间的花费我们可以求出来

现在，我们把人和花费都抽象成一个一个的点，将人和房间连一条花费为步数的边

那么求出最小费用最大流即可

代码：
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <vector>
  5 #include <cmath>
  6 #include <queue>
  7 using namespace std;
  8 
  9 const int maxn = 1005;
 10 
 11 typedef pair<int, int> PII;
 12 
 13 char a[maxn][maxn];
 14 vector<PII>v[2];
 15 const int INF = 1000000000;
 16 
 17 struct Edge {
 18     int from, to, cap, flow, cost;
 19 };
 20 
 21 struct MCMF
 22 {
 23     int n, m, s, t;
 24     vector<Edge> edges;
 25     vector<int> G[maxn];
 26 
 27     int inq[maxn];
 28     int d[maxn];
 29     int p[maxn];
 30     int a[maxn];
 31 
 32     void init(int n) {
 33         this -> n = n;
 34         for(int i = 0; i < n; i++) G[i].clear();
 35         edges.clear();
 36     }
 37 
 38     void AddEdge(int from, int to, int cap, int cost) {
 39         edges.push_back((Edge) { from, to, cap, 0, cost } );
 40         edges.push_back((Edge) { to, from, 0, 0, -cost } );
 41         m = edges.size();
 42         G[from].push_back(m - 2);
 43         G[to].push_back(m - 1);
 44     }
 45 
 46     bool BellmanFord(int s, int t, int &flow, int &cost) {
 47         for(int i = 0; i < n; i++) d[i] = INF;
 48         memset(inq, 0, sizeof(inq) );
 49         d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
 50         queue<int> Q;
 51         Q.push(s);
 52         while(!Q.empty()) {
 53             int u = Q.front(); Q.pop();
 54             inq[u] = 0;
 55             for(int i = 0; i < G[u].size(); i++) {
 56                 Edge &e = edges[G[u][i]];
 57                 if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
 58                     d[e.to] = d[u] + e.cost;
 59                     p[e.to] = G[u][i];
 60                     a[e.to] = min(a[u], e.cap - e.flow);
 61                     if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
 62                 }
 63             }
 64         }
 65 
 66         if(d[t] == INF) return false;
 67         flow += a[t];
 68         cost += d[t] * a[t];
 69         int u = t;
 70         while(u != s) {
 71             edges[p[u]].flow += a[t];
 72             edges[p[u] ^ 1].flow -= a[t];
 73             u = edges[p[u]].from;
 74         }
 75         return true;
 76     }
 77 
 78     int MinCost(int s, int t) {
 79         //this -> s = s; this -> t = t;
 80         int flow = 0, cost = 0;
 81         while(BellmanFord(s, t, flow, cost)){};
 82         return cost;
 83     }
 84 };
 85 
 86 MCMF g;
 87 
 88 int get_dist(PII p1, PII p2) {
 89     return fabs(p1.first - p2.first) + fabs(p1.second - p2.second);
 90 }
 91 
 92 int main() {
 93     int n, m;
 94     //freopen("a.txt","r",stdin);
 95     while(scanf("%d %d",&n, &m)) {
 96         if(n == 0 && m == 0)
 97         break;
 98         //puts("*");
 99         int m_num = 0, H_num = 0;
100         for(int i = 0; i < 2; i++) {
101             v[i].clear();
102         }
103         g.init(n * m);//节点个数  尽量大一点
104         for(int i = 0; i < n; i++) {
105             for(int j = 0; j < m; j++) {
106                 scanf("\n%c",&a[i][j]);//
107                 if(a[i][j] == ‘m‘) {
108                     m_num++;
109                     v[0].push_back(make_pair(i, j));//用vector把所有的人和房间都储存下来
110                 }
111                 else if(a[i][j] == ‘H‘) {
112                     H_num++;
113                     v[1].push_back(make_pair(i, j));
114                 }
115             }
116         }
117         int s = 0; int t = m_num + H_num + 1;
118         //printf("%d %d   %d %d\n",m_num, H_num, v[0].size(), v[1].size());
119         for(int i = 1; i <= m_num; i++) {
120             g.AddEdge(s, i, 1, 0);//源点跟人的容量为1 花费为 0  因为 只能输送一个人
121         }
122         for(int i = m_num + 1; i <= m_num + H_num; i++) {
123             g.AddEdge(i, t, 1, 0);//同理
124         }
125 
126         for(int i = 0; i < v[0].size(); i++) {
127             for(int j = 0; j < v[1].size(); j++) {
128                 g.AddEdge(i + 1, m_num + j + 1, 1, get_dist(v[0][i], v[1][j]));//容量为1 花费为步数
129             }
130         }
131         printf("%d\n",g.MinCost(s, t)) ;
132     }
133     return 0;
134 }
View Code

 


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