POJ 2195 Going Home

Going Home

Time Limit: 1000ms
Memory Limit: 65536KB
This problem will be judged on PKU. Original ID: 2195
64-bit integer IO format: %lld      Java class name: Main
 
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
 
解题：bfs求得每个人到每所房子的距离，然后建图做一次最小费用最大流
 
  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <queue>
  5 #define pii pair<int,int>
  6 #define INF 0x3f3f3f3f
  7 using namespace std;
  8 const int maxn = 200;
  9 struct arc {
 10     int to,flow,cost,next;
 11     arc(int x = 0,int y = 0,int z = 0,int nxt = -1) {
 12         to = x;
 13         flow = y;
 14         cost = z;
 15         next = nxt;
 16     }
 17 } e[maxn*maxn];
 18 int head[maxn*10],d[maxn*10],p[maxn*10],n,m;
 19 char mp[maxn][maxn];
 20 bool in[maxn*10];
 21 int hs[maxn][maxn],pid,hid,S,T,tot;
 22 void add(int u,int v,int flow,int cost) {
 23     e[tot] = arc(v,flow,cost,head[u]);
 24     head[u] = tot++;
 25     e[tot] = arc(u,0,-cost,head[v]);
 26     head[v] = tot++;
 27 }
 28 bool isIn(int x,int y) {
 29     return x >= 0 && x < n && y >= 0 && y < m;
 30 }
 31 void bfs(int x,int y) {
 32     queue< pii >q;
 33     q.push(make_pair(x,y));
 34     int ds[maxn][maxn];
 35     memset(ds,-1,sizeof(ds));
 36     ds[x][y] = 0;
 37     static const int dir[4][2] = {-1,0,1,0,0,-1,0,1};
 38     while(!q.empty()) {
 39         pii now = q.front();
 40         q.pop();
 41         for(int i = 0; i < 4; ++i) {
 42             int tx = now.first + dir[i][0];
 43             int ty = now.second + dir[i][1];
 44             if(isIn(tx,ty) && ds[tx][ty] == -1) {
 45                 ds[tx][ty] = ds[now.first][now.second] + 1;
 46                 q.push(make_pair(tx,ty));
 47                 if(mp[tx][ty] == 'H')
 48                     add(hs[x][y],pid+hs[tx][ty],1,ds[tx][ty]);
 49             }
 50         }
 51     }
 52 }
 53 bool spfa() {
 54     queue<int>q;
 55     for(int i = 0; i <= T; ++i) {
 56         d[i] = INF;
 57         in[i] = false;
 58         p[i] = -1;
 59     }
 60     d[S] = 0;
 61     q.push(S);
 62     while(!q.empty()) {
 63         int u = q.front();
 64         q.pop();
 65         in[u] = false;
 66         for(int i = head[u]; ~i; i = e[i].next) {
 67             if(e[i].flow && d[e[i].to] > d[u] + e[i].cost) {
 68                 d[e[i].to] = d[u] + e[i].cost;
 69                 p[e[i].to] = i;
 70                 if(!in[e[i].to]) {
 71                     in[e[i].to] = true;
 72                     q.push(e[i].to);
 73                 }
 74             }
 75         }
 76     }
 77     return p[T] > -1;
 78 }
 79 
 80 int solve() {
 81     int ans = 0;
 82     while(spfa()) {
 83         int minF = INF;
 84         for(int i = p[T]; ~i; i = p[e[i^1].to])
 85             minF = min(minF,e[i].flow);
 86         for(int i = p[T]; ~i; i = p[e[i^1].to]) {
 87             e[i].flow -= minF;
 88             e[i^1].flow += minF;
 89         }
 90         ans += d[T]*minF;
 91     }
 92     return ans;
 93 }
 94 int main() {
 95     while(scanf("%d %d",&n,&m),n||m) {
 96         memset(head,-1,sizeof(head));
 97         hid = pid = 0;
 98         for(int i = tot = 0; i < n; ++i) {
 99             scanf("%s",mp[i]);
100             for(int j = 0; j < m; ++j)
101                 if(mp[i][j] == 'H') hs[i][j] = hid++;
102                 else if(mp[i][j] == 'm') hs[i][j] = pid++;
103         }
104         S = hid + pid;
105         T = S + 1;
106         for(int i = 0; i < n; ++i)
107             for(int j = 0; j < m; ++j)
108                 if(mp[i][j] == 'm') {
109                     bfs(i,j);
110                     add(S,hs[i][j],1,0);
111                 } else if(mp[i][j] == 'H') add(hs[i][j]+pid,T,1,0);
112         printf("%d\n",solve());
113     }
114     return 0;
115 }
View Code

 

posted @ 2015-03-14 21:31  狂徒归来  阅读(185)  评论(0)  编辑  收藏  举报