HDOJ-1052 Tian Ji -- The Horse Racing

：根据题目的描述，问题可以转化为“找到最大权值的完美二分匹配”。但又说：杀鸡焉用宰牛刀。暗示了有更简单的方法。

：个人认为用到了贪心的思想“用最差的马搏最好的马”。

View Code 
 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <stdlib.h>
 4 #define NL 1100
 5 
 6 int tian[NL], king[NL];
 7 
 8 int cmp(const void *a, const void *b)
 9 {
10     return *(int*)a - *(int*)b;
11 }
12 
13 int main()
14 {
15     int n;
16     while (scanf("%d", &n) != EOF && n) {
17         int i, j;
18         for (i=0; i<n; i++) {
19             scanf("%d", &tian[i]);
20         }
21         for (i=0; i<n; i++) {
22             scanf("%d", &king[i]);
23         }
24         qsort(tian, n, sizeof(tian[i]), cmp);
25         qsort(king, n, sizeof(king[i]), cmp);
26 /*        for (i=0; i<n; i++) {
27             printf("%d\n", king[i]);
28         }            */
29         int t1, t2, k1, k2, win=0;
30         t1 = k1 =  0;
31         t2 = k2 = n-1;
32         while (t1 <= t2 && k1 <= k2) {
33             if (tian[t1] > king[k1]) {
34                 ++t1;
35                 ++k1;
36                 ++win;
37             }else if (tian[t2] > king[k2]) {
38                 --t2;
39                 --k2;
40                 ++win;
41             }else if (tian[t1] < king[k1]) {
42                 ++t1;
43                 --k2;
44                 --win;
45             }else if (tian[t2] < king[k2]) {
46                 ++t1;
47                 --k2;
48                 --win;
49             }else if (tian[t1] < king[k2]) {
50                 ++t1;
51                 --k2;
52                 --win;
53             }else if (tian[t1] > king[k2]){
54                 ++t1;
55                 --k2;
56                 ++win;                
57             }else {
58                 ++t1;
59                 --k2;
60             }
61         }
62 //        printf("t = %d\n", t);
63         printf("%d\n", 200*win);
64     }
65     return 0;
66 }
67 /*
68 4
69 1 2 3 4
70 2 3 4 1
71 3
72 1 1 1
73 1 1 1
74 3
75 90 80 70
76 90 89 79
77 */

 

posted @ 2012-10-13 17:34  superbin  阅读(323)  评论(0)  编辑  收藏  举报