ZCMU-1040：二哥的困惑Ⅲ（贪心）

最新推荐文章于 2022-07-18 19:15:57 发布

waterboy_cj

最新推荐文章于 2022-07-18 19:15:57 发布

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本文介绍了一个特殊卡片游戏的策略分析，玩家通过出牌竞争胜利。文章详细解释了游戏规则，并提供了一种有效的算法来确定玩家至少能赢得多少轮比赛。

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1040: 二哥的困惑 Ⅲ

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 216 Solved: 84
[Submit][Status][Web Board]

Description

Suppose there are M people, including you, playing a special card game. At the beginning, each player receives N cards. The pip of a card is a positive integer which is at most N*M. And there are no two cards with the same pip. During a round, each player chooses one card to compare with others. The player whose card with the biggest pip wins the round, and then the next round begins. After N rounds, when all the cards of each player have been chosen, the player who has won the most rounds is the winner of the game.

Given your cards received at the beginning, write a program to tell the maximal number of rounds that you may at least win during the whole game.

Input

The input consists of several test cases. The first line of each case contains two integers m (2<=m<=20) and n (1<=n<=50), representing the number of players and the number of cards each player receives at the beginning of the game, respectively. This followed by a line with n positive integers, representing the pips of cards you received at the beginning. Then a blank line follows to separate the cases.

The input is terminated by a line with two zeros.

Output

For each test case, output a line consisting of the test case number followed by the number of rounds you will at least win during the game.

Sample Input

2 5

1 7 2 10 9

6 11

62 63 54 66 65 61 57 56 50 53 48

0 0

Sample Output

Case 1: 2

Case 2: 4

HINT

Source

【解析】

题意：m个玩家,给你n张牌，所有牌的编号为1-m*n。给你你所拥有的牌，每人每局出一张牌，问，你至少能赢几把。

贪心很明显了。

我的做法是，从m*n开始遍历，代表你每次都出最大的牌，如果最大的你有了，那你肯定赢了，now++,然后及时更新ans。如果当前最大的你没有，那么人家就可能比你大。所以now--。

#include <bits/stdc++.h>
using namespace std;
int main()
{
	int n, m, num[55], flag[1010], ca = 0;
	while (scanf("%d%d", &m, &n))
	{
		if (!m&&!n)break;
		memset(flag, 0, sizeof(flag));
		int ans = 0, now = 0;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &num[i]);
			flag[num[i]] = 1;
		}
		for (int i = m * n; i > 0; i--)
		{
			if (flag[i])
			{
				now++;
				if (now > ans)
					ans = now;
			}
			else
				now--;
		}
		printf("Case %d: %d\n", ++ca, ans);
	}

	return 0;
}