poj1323(贪心)

http://poj.org/problem?id=1323

Game Prediction

Time Limit: 1000MSMemory Limit: 10000KTotal Submissions: 8292Accepted: 3976

Description

Suppose there are M people,including you, playing a special card game. At the beginning, eachplayer receives N cards. The pip of a card is a positive integerwhich 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 withothers. The player whose card with the biggest pip wins the round,and then the next round begins. After N rounds, when all the cardsof each player have been chosen, the player who has won the mostrounds is the winner of the game.



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

Input

The input consists of severaltest cases. The first line of each case contains two integers m(2?20) and n (1?50), representing the number of players and thenumber 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. Thena blank line follows to separate the cases.

The input is terminated by a line with two zeros.

Output

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

Sample Input

2 51 7 2 10 96 1162 63 54 66 65 61 57 56 50 53 480 0

Sample Output

Case 1: 2

Case 2: 4

Source

Beijing 2002

题意：这题真的理解了很久啊一直不理解。At thebeginning, each player receives N cards. The pip of a card is apositive integer which is at most N*M. And there are no two cardswith the same pip.

这句是关键就是每个人有N张卡片，且每个人的卡片都不一样。(刚开始一直理解成这些卡片是相同的)嗨，理解的太偏。这些牌的序号是从1——N*M的

题意：n个人在玩牌，每个人有m张牌，于是就有n*m张牌（每张牌都有一个值，介于1到n*m之间，不重复），然后进行m轮游戏，每轮每个人都出一张牌，牌最大的那个人就赢了，然后给出n和m，以及你的m张牌，问你最多能赢几轮？

思路：将除了你自己手中牌的数字的其他最大的m个数字的牌给予另一个人，即给予该人一副最优的牌，接下来就是判断你和这个人玩这个游戏，你至少能获胜几轮。

参考资料http://blog.csdn.net/night146/article/details/6045172

太谢谢了。。。

网上还有一种很普遍的做法，但是我还有个点不理解。。

http://blog.csdn.net/night146/article/details/6045172

discuss里面有一种和我想的差不多。。也就是和最上面的那位其实差不多http://poj.org/showmessage?message_id=148827

 

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int me[60],other[60];
int used[10050];
int n,m;
int Case=0,ans;
int cmp(const void *a,const void *b)
{
    return *(int *)a - *(int *)b;

}
int main()
{
 int i,j;
 while(scanf("%d%d",&n,&m)&& n!=0 )
 {

  ans=0;
  Case++;
  memset(used,0,sizeof(used));
       for(i=0;i<m;i++)
  {
   scanf("%d",&me[i]);
   used[me[i]]=1;
  }
  for(i=n*m,j=0;j<m;i--)//给予某个人一副最优的牌
  {
   if(!used[i])
   {
                 other[j]=i;
     used[other[j]]=1;
     j++;
   }
  }
       qsort(me,m,sizeof(me[0]),cmp);
  qsort(other,m,sizeof(other[0]),cmp);
  for(i=m-1;i>=0;i--)//你从大出道小，而另一个人只需要出他手中的比你大的牌中的最小的一张，即该轮你是输的
  {
            for(j=0;j<m;j++)
    {
    if(other[j]>me[i] &&used[other[j]])
    {
     used[other[j]]=0;
     break;
    }
    }
  }
  for(i=0;i<m;i++)//判断那一个人输的次数，也就是你赢的次数
   if(used[other[i]])
    ans++;
   printf("Case%d: %d\n",Case,ans);
 }
 return 0;
}