HDU 1338 Game Prediction【贪心】

最新推荐文章于 2022-11-13 20:15:43 发布

原创最新推荐文章于 2022-11-13 20:15:43 发布 · 450 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#acm-icpc #HDU #1338 #贪心

算法专栏收录该内容

45 篇文章

订阅专栏

本文介绍了一种特殊纸牌游戏的预测算法，旨在确定玩家在比赛中能够至少获胜的局数。通过分析玩家手上的牌和对手可能持有的牌，该算法提供了一个有效的策略来最大化玩家的胜率。

慧航空AI大赛”（报名中...）

Game Prediction

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1613 Accepted Submission(s): 915

Problem 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

题意：有m个人，每人得到n张牌，牌的号码是1到n*m之间的任意n个数，每张牌都只有一张，问我至少能赢多少局。

分析：

可以转化为你最多输max次，那么至少赢n-max次而最多输max次，则是对方最多赢max次，则用对方的最小的牌去依次比较你手中的牌（按照升序排），如果找到有比它小的，则对方赢一次依次循环直到遍历完对方的牌。

#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<string>
#include<cstring>
#include<cstdio>
const int maxn=100005;
const int INF=0x3f3f3f3f;
typedef long long LL;
using namespace std;
int a[maxn],vis[maxn];
int used[maxn];
int main()
{
//	freopen("E:\\ACM\\test.txt","r",stdin);
	int N,M;
	int t=1;
	while(cin>>M>>N,M+N)
	{
		memset(vis,0,sizeof(vis));
		memset(used,0,sizeof(used));
		int ans=0; //我最多可能输的次数 
		for(int i=0;i<N;i++) 
		{
			cin>>a[i];
			vis[a[i]]=1; //vis[i]=0表示对手有的牌 
		}
		sort(a,a+N);
		for(int i=1;i<=M*N;i++)
		{
			if(!vis[i])
			for(int j=0;j<N;j++)
			{
				if(!used[j])
				{
              		if(a[j]<i) //我的牌中最小的比对手最小的还小，则我输 
					{
						++ans; 
						used[j]=1; //表示我的这张牌已经比较过 
						break; //注意这里要跳出循环，比较对手下一张牌 
					} 					
				}
			}
		}
		printf("Case %d: %d\n",t++,N-ans);
		
	}
	
	return 0;
}