Going Home
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3299 Accepted Submission(s): 1674
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
Source
题目意思：
给一个n*m的地图，‘m‘表示人，‘H‘表示房子，人每次能向上下左右走到邻近的单位，每次走一次总费用+1，每个房子只能容纳一个人，求当所有人走到房子后总费用最小为多少。
思路：
人放左边，房子放右边，之间的边权值表示人x[i]走到房子y[j]处花费的费用。权值弄成负数，KM找最佳匹配，答案在负一下即可。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 #include <cmath>
8 #include <set>
9 using namespace std;
10
11 #define N 105
12 #define inf 999999999
13
14 int max(int x,int y){return x>y?x:y;}
15 int min(int x,int y){return x<y?x:y;}
16 int abs(int x,int y){return x<0?-x:x;}
17
18 struct KM {
19 int n, m;
20 int g[N][N];
21 int Lx[N], Ly[N], slack[N];
22 int left[N], right[N];
23 bool S[N], T[N];
24
25 void init(int n, int m) {
26 this->n = n;
27 this->m = m;
28 memset(g, 0, sizeof(g));
29 }
30
31 void add_Edge(int u, int v, int val) {
32 g[u][v] += val;
33 }
34
35 bool dfs(int i) {
36 S[i] = true;
37 for (int j = 0; j < m; j++) {
38 if (T[j]) continue;
39 int tmp = Lx[i] + Ly[j] - g[i][j];
40 if (!tmp) {
41 T[j] = true;
42 if (left[j] == -1 || dfs(left[j])) {
43 left[j] = i;
44 right[i] = j;
45 return true;
46 }
47 } else slack[j] = min(slack[j], tmp);
48 }
49 return false;
50 }
51
52 void update() {
53 int a = inf;
54 for (int i = 0; i < m; i++)
55 if (!T[i]) a = min(a, slack[i]);
56 for (int i = 0; i < n; i++)
57 if (S[i]) Lx[i] -= a;
58 for (int i = 0; i < m; i++)
59 if (T[i]) Ly[i] += a;
60 }
61
62 int km() {
63 memset(left, -1, sizeof(left));
64 memset(right, -1, sizeof(right));
65 memset(Ly, 0, sizeof(Ly));
66 for (int i = 0; i < n; i++) {
67 Lx[i] = -inf;
68 for (int j = 0; j < m; j++)
69 Lx[i] = max(Lx[i], g[i][j]);
70 }
71 for (int i = 0; i < n; i++) {
72 for (int j = 0; j < m; j++) slack[j] = inf;
73 while (1) {
74 memset(S, false, sizeof(S));
75 memset(T, false, sizeof(T));
76 if (dfs(i)) break;
77 else update();
78 }
79 }
80 int ans = 0;
81 for (int i = 0; i < n; i++) {
82 ans += g[i][right[i]];
83 }
84 return ans;
85 }
86 }kmm;
87
88 int n, m;
89 char map[N][N];
90 int g[N][N];
91
92 struct node{
93 int x, y;
94 node(){}
95 node(int a,int b){
96 x=a;
97 y=b;
98 }
99 }a[N], b[N];
100
101 main()
102 {
103 int i, j, k;
104 int nx, ny;
105 while(scanf("%d %d",&n,&m)==2){
106 if(n==0&&m==0) break;
107
108 for(i=0;i<n;i++) scanf("%s",map[i]);
109 nx=ny=0;
110 memset(g,0,sizeof(g));
111 for(i=0;i<n;i++){
112 for(j=0;j<m;j++){
113 if(map[i][j]==‘m‘) a[nx++]=node(i,j);
114 if(map[i][j]==‘H‘) b[ny++]=node(i,j);
115 }
116 }
117 kmm.init(nx,ny);
118 for(i=0;i<nx;i++){
119 for(j=0;j<ny;j++){
120 kmm.add_Edge(i,j,-(abs(a[i].x-b[j].x)+abs(a[i].y-b[j].y)));
121 }
122 }
123 printf("%d\n",-kmm.km());
124 }
125 }