POJ 2912 Rochambeau

题目描述：
　　石头剪子布游戏，总共N个小孩，把N - 1个分成3组，每组的小孩每次只能出固定的。一人为裁判，每次可以随意出。进行M场比赛，结果用X>Y、X<Y、X=Y表示。问经过几轮能推断出裁判的ID。

题目类型：并查集

解题思路：　　　　　　
（1）依次假设每个小孩为裁判，遍历每个比赛（遇到有该小孩的则跳过），用并查集求出产生矛盾的轮数。若有多个人没产生矛盾，输出Can not determine。若所有人都产生矛盾，输出Impossible。若只有一人没产生矛盾，则其为裁判，其余小孩产生矛盾的最大轮数是推断出裁判的最大轮数。
（2）节点与根节点的关系：根为0，若根胜则节点值为2，若根负则节点值为1，若根平则节点值为0。合并两个树时，讲r2合并到r1，即root[r2] = r1。更新关系：relation[r2] = (relation[s1] - relation[s2] + relation[r1] + r + 3) % 3。判断两节点是否矛盾：(relation[s1] + r) % 3 == relation[s2]，相等说明不矛盾。

运行结果：500MS + 200K

View Code

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include "stdio.h"
 3 #include "string.h"
 4 
 5 int N, M, root[501], relation[501];
 6 struct ROUND { int s1, s2, r; } round[10001];
 7 
 8 int find(int s)
 9 {
10     if (root[s] != s) 
11     {
12         int r = root[s];
13         root[s] = find(root[s]);
14         relation[s] = (relation[s] + relation[r]) % 3;
15     }
16     return root[s];
17 }
18 
19 bool add(int s1, int s2, int r)
20 {
21     int r1 = find(s1);
22     int r2 = find(s2);
23 
24     if (r1 != r2) root[r2] = r1, relation[r2] = (relation[s1] - relation[s2] + relation[r1] + r + 3) % 3;
25     else if ((relation[s1] + r) % 3 != relation[s2]) return false;
26     return true;
27 }
28 
29 void solve()
30 {
31     int judge_sum = 0, judge_num = -1, max_round = -1, k;
32     for (int i = 0; i < N; ++i)
33     {
34         for (k = 0; k < N; ++k) root[k] = k;
35         memset(relation, 0, sizeof(int) * N);
36         for (k = 0; k < M; ++k)
37         {
38             if (round[k].s1 == i || round[k].s2 == i) continue;
39             if (!add(round[k].s1, round[k].s2, round[k].r))
40             {
41                 if (k > max_round) max_round = k;
42                 break;
43             }
44         }
45         if (k == M) ++judge_sum, judge_num = i;
46     }
47 
48     if (judge_sum == 0) printf("Impossible\n");
49     else if (judge_sum > 1) printf("Can not determine\n");
50     else printf("Player %d can be determined to be the judge after %d lines\n", 
51         judge_num, max_round + 1);
52 }
53 
54 int main()
55 {
56     while (scanf("%d%d", &N, &M) != -1)
57     {
58         for (int i = 0; i < M; ++i)
59         {
60             scanf("%d", &round[i].s1);
61             char ch = getchar();
62             round[i].r = (ch == '>' ? 2 : (ch == '<' ? 1 : 0)); 
63             scanf("%d", &round[i].s2);
64         }
65         solve();
66     }
67     return 0;
68 }

posted on 2011-03-02 23:46 theParanoia 阅读(196) 评论(0) 收藏举报