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Going Home
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6641 Accepted Submission(s): 3491
Problem 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

C/C++：

  1 #include <map>
  2 #include <queue>
  3 #include <cmath>
  4 #include <vector>
  5 #include <string>
  6 #include <cstdio>
  7 #include <cstring>
  8 #include <climits>
  9 #include <iostream>
 10 #include <algorithm>
 11 #define
 
 INF 0x3f3f3f3f
 12 using namespace std;
 13 const int my_max = 110, my_max_hm = 5010;
 14 
 15 int my_map[my_max][my_max], n, m, N, my_line[my_max_hm]
 16     , my_lx[my_max_hm], my_ly[my_max_hm], my_slack[my_max_hm]
 17     , my_bookx[my_max_hm], my_booky[my_max_hm];
 18 
 19 struct node
 20 {
 21     int
 
 x, y;
 22 } my_home[my_max_hm], my_men[my_max_hm];
 23 
 24 bool my_dfs(int x)
 25 {
 26     my_bookx[x] = 1;
 27     for (int i = 1; i <= N; ++ i)
 28     {
 29         if (my_booky[i]) continue;
 30         int temp = my_lx[x] + my_ly[i] - my_map[x][i];
 31         if (temp == 0)
 32         {
 33             my_booky[i] = 1;
 34             if (!my_line[i] || my_dfs(my_line[i]))
 35             {
 36                 my_line[i] = x;
 37                 return true;
 38             }
 39         }
 40         else if (my_slack[i] > temp)
 41             my_slack[i] = temp;
 42     }
 43     return false;
 44 }
 45 
 46 int my_km()
 47 {
 48     memset(my_line, 0, sizeof(my_line));
 49     memset(my_ly, 0, sizeof(my_ly));
 50     for (int i = 1; i <= N; ++ i)
 51     {
 52         my_lx[i] = -INF;
 53         for (int j = 1; j <= N; ++ j)
 54             if (my_lx[i] < my_map[i][j])
 55                 my_lx[i] = my_map[i][j];
 56     }
 57 
 58     for (int i = 1; i <= N; ++ i)
 59     {
 60         for (int j = 1; j <= N; ++ j)
 61             my_slack[j] = INF;
 62         while (1)
 63         {
 64             memset(my_bookx, 0, sizeof(my_bookx));
 65             memset(my_booky, 0, sizeof(my_booky));
 66 
 67             if (my_dfs(i)) break;
 68             int my_temp_min = INF;
 69             for (int j = 1; j <= N; ++ j)
 70                 if (!my_booky[j] && my_slack[j] < my_temp_min)
 71                     my_temp_min = my_slack[j];
 72 
 73             for (int j = 1; j <= N; ++ j)
 74                 if (my_bookx[j]) my_lx[j] -= my_temp_min;
 75             for (int j = 1; j <= N; ++ j)
 76                 if (my_booky[j]) my_ly[j] += my_temp_min;
 77                 else my_slack[j] -= my_temp_min;
 78         }
 79     }
 80     int my_ans = 0;
 81     for (int i = 1; i <= N; ++ i)
 82         my_ans += my_map[my_line[i]][i];
 83     return my_ans;
 84 }
 85 
 86 int main()
 87 {
 88     while (scanf("%d%d", &n, &m), n || m)
 89     {
 90         int my_cnt_h = 0, my_cnt_m = 0;
 91         memset(my_map, 0, sizeof(my_map));
 92         getchar();
 93         for (int i = 1; i <= n; ++ i)
 94         {
 95             char my_s[my_max];
 96             scanf("%s", my_s);
 97             for (int j = 0; j < m; ++ j)
 98             {
 99                 if (my_s[j] == ‘H‘)
100                 {
101                     my_home[my_cnt_h].x = i;
102                     my_home[my_cnt_h].y = j;
103                     my_cnt_h ++;
104                 }
105                 else if (my_s[j] == ‘m‘)
106                 {
107                     my_men[my_cnt_m].x = i;
108                     my_men[my_cnt_m].y = j;
109                     my_cnt_m ++;
110                 }
111             }
112         }
113 
114         N = my_cnt_h;
115         for (int i = 1; i <= N; ++ i)
116         {
117             for (int j = 1; j <= N; ++ j)
118             {
119                 my_map[i][j] += abs(my_men[i - 1].x - my_home[j - 1].x);
120                 my_map[i][j] += abs(my_men[i - 1].y - my_home[j - 1].y);
121                 my_map[i][j] *= -1;
122             }
123         }
124         printf("%d\n", -1 * my_km());
125     }
126     return 0;
127 }

hdu 1533 Going Home (Dinic, 1)