ZCMU 1040: 二哥的困惑 Ⅲ

最新推荐文章于 2022-03-26 12:35:32 发布

不吃辣的李叔叔

最新推荐文章于 2022-03-26 12:35:32 发布

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分类专栏：贪心算法 ZCMU

本文链接：https://blog.youkuaiyun.com/qq_39315193/article/details/81570227

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本文介绍了一种特殊卡牌游戏的最优策略，玩家需在每轮比赛中选择一张卡牌进行比拼，目标是在限定轮次中赢得尽可能多的回合。通过将问题抽象为一种策略选择问题，并采用类似田忌赛马的策略来确定至少可以赢得的回合数。

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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

【分析】这题和田忌赛马https://blog.youkuaiyun.com/qq_39315193/article/details/81563882解题思路相同，更简单

虽然是多人游戏，如果把其他人看成一队，全针对自己，这样就能保证我赢的是最少的。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
int a[1005], b[1005];
bool cmp(int a, int b)
{
    return a > b;
}
int main()
{
    int m,n ;
    int count = 0;
    while(scanf("%d%d",&m,&n))
    {
        if(m==0&&n==0)break;
        count++;
        memset(a , 0 ,sizeof(a));
        int t = 0,win = 0;;
        for(int i =0;i<n;i++)
            scanf("%d",&a[i]);
        sort(a, a + n, cmp);//自己牌a[]降序排列
        for(int i = n*m,j=0;i>0;i--)
        {
            if(i==a[t])
            {
                t++;
                continue;
            }
            b[j] = i;
            j++;
        }//对面牌b[]降序排列

        for(int i = 0,j =0;i<n;i++)//我从大往小出牌
        {
            if(b[j]>a[i])//对手最大的牌能大于我的牌，对面就出最大的（换个理解就是对面用最大的牌耗我最大的牌）
                j++;                                               
            else
                win++;//对面最大的牌小于我的最大的牌，对面就出最小的，j不变表示最大的牌留着
                      //因为不存在两张牌相同，所以不必考虑最小牌耗我的最小牌的情况（当对手最小牌大于我的最小牌）
        }
        printf("Case %d: %d\n" ,count ,win);
    }
    return 0;
}

这篇解题思路很好https://blog.youkuaiyun.com/abcjennifer/article/details/5405097