#1154 :Spring Outing

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我是靠谱客的博主柔弱宝贝，最近开发中收集的这篇文章主要介绍#1154 :Spring Outing，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

时间限制: 20000ms

单点时限: 1000ms

内存限制: 256MB

描述

You class are planning for a spring outing. N people are voting for a destination out of K candidate places.

The voting progress is below:

First the class vote for the first candidate place. If more than half of the class agreed on the place, the place is selected. The voting ends.

Otherwise they vote for the second candidate place. If more than half of the class agreed on the place, the place is selected. The voting ends.

Otherwise they vote for the third candidate place in the same way and go on.

If no place is selected at last there will be no spring outing and everybody stays at home.

Before the voting, the Chief Entertainment Officer did a survey, found out every one's preference which can be represented as a permutation of 0, 1, ... K. (0 is for staying at home.) For example, when K=3, preference "1, 0, 2, 3" means that the first place is his first choice, staying at home is the second choice, the second place is the third choice and the third place is the last choice.

The Chief Entertainment Officer sends the survey results to the class. So everybody knows the others' preferences. Everybody wants his more prefered place to be selected. And they are very smart, they always choose the optimal strategy in the voting progress to achieve his goal.

Can you predict which place will be selected?

输入

The first line contains two integers, N and K, the number of people in your class and the number of candidate places.

The next N lines each contain a permutation of 0~K, representing someone's preference.

For 40% of the data, 1 <= N, K <= 10

For 100% of the data, 1 <= N, K <= 1000

输出

Output the selected place. Or "otaku" without quotes if no place is selected.

样例提示

In the sample case, if the second peoson vote against the first place, no place would be selected finally because the first person must vote against the second place for his own interest. Considering staying at home is a worse choice than the first place, the second person's optimal strategy is voting for the first place. So the first place will be selected.

样例输入

2 2
1 0 2
2 1 0

样例输出

参考了该链接思路 http://www.cnblogs.com/Patt/p/4678756.html

用 java 重写了一下

import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;

/**
 * 
 * @author
 *逆推，最后一个城市投票时，只有两种可能性，选最后一个还是在家，只要这个城市的优先级比 在家高，那么肯定会投这个城市
 *这样最后一个城市的投票结果就预先知道了R(k)， 倒数第二个城市投票的时候，只有 该城市和 R(k)两种可能，所以倒数第二个城
 *市的结果也知道了，依次类推
 */
public class Main {

	public static void main(String[] args) throws IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(new FileInputStream("F:/in.txt")));
		String[] num = bf.readLine().split(" ");
				
		int N = Integer.valueOf(num[0]);
		int K = Integer.valueOf(num[1]);
		// place and weight
		Map<Integer, Integer>[] pref = new HashMap[N];
		for(int i = 0; i < N; i++)
		{
			pref[i] = new HashMap<Integer, Integer>();
		}
		for(int i = 0; i < N; i++)
		{
			String[] place = bf.readLine().split(" ");
			for(int j = 0; j <= K; j++)
			{
				pref[i].put(Integer.valueOf(place[j]), K - j);
			}
		}
		int preR = 0;
		//倒推
		for(int pla = K; pla > 0; pla--)
		{
			//N 个人投票
			int voteNum = 0;
			for(int people = 0; people < N; people++)
			{
				if(pref[people].get(pla) > pref[people].get(preR))
				{
					voteNum++;
				}
			}
			if(voteNum >= N / 2 + 1)
			{
				preR = pla;
			}
		}
		if(preR == 0)
		{
			System.out.println("otaku");
		}
		else
		{
			System.out.println(preR);
		}
		

	}

}