（第六场）Singing Contest 【模拟】

题目链接：https://www.nowcoder.com/acm/contest/144/A

标题：A、Singing Contest

| 时间限制：1 秒 | 内存限制：256M

Jigglypuff is holding a singing contest. There are 2n singers indexed from 1 to 2n participating in the contest. The rule of this contest is like the knockout match. That is, in the first round, singer 1 competes with singer 2, singer 3 competes with singer 4 and so on; in the second round, the winner of singer 1 and singer 2 competes with the winner of singer 3 and singer 4 and so on. There are n rounds in total. Each singer has prepared n songs before the contest. Each song has a unique pleasantness. In each round, a singer should sing a song among the songs he prepared. In order not to disappoint the audience, one song cannot be performed more than once. The singer who sings the song with higher pleasantness wins. Now all the singers know the pleasantness of songs prepared by all the others. Everyone wants to win as many rounds as he can. Assuming that singers choose their song optimally, Jigglypuff wants to know which singer will win the contest?
输入描述: The input starts with one line containing exactly one integer t which is the number of test cases. (1 ≤ t ≤ 10)

For each test case, the first line contains exactly one integer n where 2n is the number of singers. (1 ≤ n ≤ 14)

Each of the next 2n lines contains n integers where aij is the pleasantness of the j-th song of the ith singer. It is guaranteed that all these 2nx n integers are pairwise distinct. (1≤ aij ≤ 109)

输出描述: For each test case, output "Case #x: y" in one line (without quotes), where x is the test case number (starting from 1) and y is the index of the winner.

示例 1

输入

2

1

1

2

2

1 8

2 7

3 4

5 6

输出

Case #1: 2

Case #2: 4

 

题意概括：

歌唱比赛，有2^N位歌手，每位歌手准备N首歌，每首歌可以得到的分数不同，每首歌只能唱一次。1和2比，3和4比...赢了的继续比下去，问最后谁会获胜。每位歌手的歌曲得分用一个二维矩阵表示，A[ i ][ j ]表示第 i 位歌手唱第 j 首歌可以得到的分数。

官方题解：

由于每个选⼿手的策略略都是尽可能赢，所以他该认输的时候只能认输。
能赢的时候只要选权值⼤大于对⽅方最⼤大值的最⼩小值，⼤大的留留在后⾯面不不会 更更差。
直接模拟即可。

解题思路：

每次对决，遵循贪心的原则，排序之后lower_bound()可以打败对手的最小值，遍历对决可以用DFS二分一下。

 

AC code：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int MAXN = (1<<14)+15;
 5 int f[MAXN][15];
 6 int N;
 7 
 8 int dfs(int l, int r)
 9 {
10     if(r == l+1)
11     {
12         sort(f[l], f[l]+N);
13         sort(f[r], f[r]+N);
14         int a = lower_bound(f[l], f[l]+N, f[r][N-1])-f[l];
15         int b = lower_bound(f[r], f[r]+N, f[l][N-1])-f[r];
16         if(a == N)
17         {
18             f[r][b] = 0;
19             return r;
20         }
21         else
22         {
23             f[l][a] = 0;
24             return l;
25         }
26     }
27     else
28     {
29         int mid = (l+r)>>1;
30         int x = dfs(l, mid);
31         int y = dfs(mid+1, r);
32         sort(f[x], f[x]+N);
33         sort(f[y], f[y]+N);
34         int a = lower_bound(f[x], f[x]+N, f[y][N-1])-f[x];
35         int b = lower_bound(f[y], f[y]+N, f[x][N-1])-f[y];
36         if(a == N)
37         {
38             f[y][b] = 0;
39             return y;
40         }
41         else
42         {
43             f[x][a] = 0;
44             return x;
45         }
46     }
47 }
48 
49 int main()
50 {
51     int T_case;
52     scanf("%d", &T_case);
53     int cnt = 0;
54     while(T_case--)
55     {
56         scanf("%d", &N);
57         for(int i = 1; i <= (1<<N); i++)
58             for(int j = 0; j < N; j++)
59         {
60             scanf("%d", &f[i][j]);
61         }
62         printf("Case #%d: %d\n", ++cnt, dfs(1, (1<<N)));
63     }
64     return 0;
65 }

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