Going Home
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17955 | Accepted: 9145 |
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扫描路径得到人和房子的坐标,相减得所费路程,然后建边最小费用流即可
1 #include<cstdio>
2 #include <queue>
3 #include <algorithm>
4 #include <assert.h>
5 #include <cstring>
6 using namespace std;
7 const int inf=0x7fffffff;
8 int n,m;
9
10 char maz[101][101];
11 int man[101][2];
12 int house[101][2];
13 int hlen,mlen;
14
15
16 const int sups=201;
17 const int supt=202;
18
19 int cost[203][203];
20 int f[203][203];
21 int e[203][203];
22 int len[203];
23
24 int d[203];
25 int pre[203];
26 bool vis[203];
27
28 queue<int >que;
29
30 int main(){
31 while(scanf("%d%d",&n,&m)==2&&n&&m){
32 input:
33 hlen=mlen=0;
34 gets(maz[0]);
35 for(int i=0;i<n;i++){
36 gets(maz[i]);
37 }
38 for(int i=0;i<n;i++){
39 for(int j=0;j<m;j++){
40 if(maz[i][j]==‘m‘){
41 man[mlen][0]=i;man[mlen++][1]=j;
42 }
43 else if(maz[i][j]==‘H‘){
44 house[hlen][0]=i;house[hlen++][1]=j;
45 }
46 }
47 }
48
49 memset(f,0,sizeof(f));
50 memset(cost,0,sizeof(cost));
51 fill(len,len+mlen,hlen+1);fill(len+mlen,len+mlen+hlen,mlen+1);len[sups]=mlen;len[supt]=hlen;
52 bulidedge:
53 for(int i=0;i<mlen;i++){
54 f[sups][i]=1;
55 e[sups][i]=i;
56 e[i][0]=sups;
57 for(int j=0;j<hlen;j++){
58 e[i][j+1]=j+mlen;
59 e[j+mlen][i+1]=i;
60 f[i][j+mlen]=1;
61 cost[i][j+mlen]=abs(man[i][0]-house[j][0])+abs(man[i][1]-house[j][1]);
62 cost[j+mlen][i]=-cost[i][j+mlen];
63 }
64 }
65 for(int j=0;j<hlen;j++){
66 e[supt][j]=j+mlen;
67 e[j+mlen][0]=supt;
68 f[j+mlen][supt]=1;
69 }
70
71 mincostflow:
72 int ans=0;
73 int flow=mlen;
74 while(flow>0){
75 fill(d,d+203,inf);
76 d[sups]=0;
77 que.push(sups);
78 while(!que.empty()){
79 int fr=que.front();que.pop();
80 vis[fr]=false;
81 for(int i=0;i<len[fr];i++){
82 int to=e[fr][i];
83 if(f[fr][to]>0&&d[to]>d[fr]+cost[fr][to]){
84 d[to]=d[fr]+cost[fr][to];
85 pre[to]=fr;
86 if(!vis[to]){
87 que.push(to);
88 vis[to]=true;
89 }
90 }
91 }
92 }
93
94 assert(d[supt]!=inf);
95
96 int sub=flow;
97 for(int i=supt;i!=sups;i=pre[i]){
98 sub=min(flow,f[pre[i]][i]);
99 }
100 flow-=sub;
101 ans+=sub*d[supt];
102 for(int i=supt;i!=sups;i=pre[i]){
103 f[pre[i]][i]-=sub;
104 f[i][pre[i]]+=sub;
105 }
106
107 }
108 printf("%d\n",ans);
109 }
110 return 0;
111 }
最小费用流 poj2195