HDU 1150 - Machine Schedule

今天熟悉了一下匈牙利算法。简单来说，匈牙利算法解决的问题就是二分图的无权完备匹配问题，关于带权完备匹配参见Kuhn–Munkres algorithm.

关于这方面的知识，主要的参考资料是ByVoid的两篇：

https://www.byvoid.com/blog/hungary/

https://www.byvoid.com/blog/match-km/

Wikipedia

http://en.wikipedia.org/wiki/Hungarian_algorithm

还有Harold W. Kuhn的一篇论文

The Hungarian Method for the assignment problem

 1 /*
 2 ID:esxgx1
 3 LANG:C++
 4 PROG:hdu1150
 5 */
 6 #include <cstdio>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 
12 #define MAXN    107
13 #define MAXM    107
14 
15 int N, M;
16 int visited[MAXM];
17 int matched[MAXM];
18 
19 
20 int e[MAXN][MAXN];
21 
22 int aug(int i)
23 {
24     for(int j=0; j<M; ++j) {
25         if (e[i][j] && !visited[j]) {
26             visited[j] = 1;
27             if (matched[j] < 0 || aug(matched[j])) {
28                 matched[j] = i;
29                 return 1;
30             }
31         }
32     }
33     return 0;
34 }
35 
36 int main(void)
37 {
38     #ifndef ONLINE_JUDGE
39     freopen("in.txt", "r", stdin);
40     #endif
41 
42     int K;
43     while(scanf("%d%d%d", &N, &M, &K) >= 3) {
44         memset(e, 0, sizeof(e));
45         for(int i=0; i<K; ++i) {
46             int _, x, y;
47             scanf("%d%d%d", &i, &x, &y);
48             if (x > 0 && y > 0) e[x][y] = 1;
49         }
50         memset(matched, -1, sizeof(matched));
51         // hungrian method
52         int m = 0;
53         for(int i=0; i<N; ++i) {
54             m += aug(i);
55             memset(visited, 0, sizeof(visited));
56         }
57         printf("%d\n", m);
58     }
59     return 0;
60 }

 

2014-09-17 02:09:32 Accepted 1150 0MS 352K 972 B G++
posted @ 2014-09-17 02:22  e0e1e  阅读(280)  评论(0)  编辑  收藏  举报