LightOJ 1400 Employment (Stable Marriage)

　　一道关于稳定婚姻的匹配题，算法的详细解释可以参考Matrix67的文章。

下面是我根据算法思路写出来的代码：

View Code

 1 int comCan[N][N];
 2 int like[N], canLV[N][N];
 3 
 4 void input(int n) {
 5     int x;
 6     REP_1(i, n) {
 7         REP_1(j, n) {
 8             scanf("%d", &x);
 9             comCan[i][j] = x - n;
10         }
11     }
12     REP_1(i, n) {
13         REP_1(j, n) {
14             scanf("%d", &x);
15             canLV[i][x] = j;
16         }
17     }
18     REP_1(i, n) like[i] = 1;
19 }
20 
21 void work(int n) {
22     int mk[N];
23     bool ok;
24     while (true) {
25         REP_1(i, n) mk[i] = 0;
26         REP_1(i, n) {
27             int can = comCan[i][like[i]];
28             if (!mk[can]) mk[can] = i;
29             else {
30                 if (canLV[can][i] < canLV[can][mk[can]]) mk[can] = i;
31             }
32         }
33         ok = true;
34         REP_1(i, n) if (!mk[i]) {
35             ok = false;
36             break;
37         }
38         if (ok) break;
39         REP_1(i, n) {
40             int can = comCan[i][like[i]];
41             if (mk[can] != i) like[i]++;
42         }
43     }
44 }
45 
46 void output(int c, int n) {
47     printf("Case %d:", c);
48     REP_1(i, n) printf(" (%d %d)", i, comCan[i][like[i]] + n);
49     puts("");
50 }
51 
52 int main() {
53     int T, n, CAS = 0;
54     scanf("%d", &T);
55     while (T-- && ~scanf("%d", &n)) {
56         input(n);
57         work(n);
58         output(++CAS, n);
59     }
60     return 0;
61 }

——written by Lyon

posted @ 2012-12-29 16:17 LyonLys 阅读(229) 评论(0) 收藏举报

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