poj 3664 Election Time

最新推荐文章于 2019-08-10 13:10:53 发布

转载最新推荐文章于 2019-08-10 13:10:53 发布 · 136 阅读

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原文链接：http://www.cnblogs.com/w0w0/archive/2011/11/23/2259864.html

在这篇博客中，作者介绍了如何通过分析牛群选举的投票数据来预测胜选者。选举分为两轮，第一轮中得票最多的K头牛进入第二轮，最终一轮得票最高的牛将成为总统。作者提供了代码实现，包括对投票数据的排序和比较，以确定胜选者。

Election Time

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5030		Accepted: 2750

Description

The cows are having their first election after overthrowing the tyrannical Farmer John, and Bessie is one of N cows (1 ≤ N ≤ 50,000) running for President. Before the election actually happens, however, Bessie wants to determine who has the best chance of winning.

The election consists of two rounds. In the first round, the K cows (1 ≤ K ≤ N) cows with the most votes advance to the second round. In the second round, the cow with the most votes becomes President.

Given that cow i expects to get A_i votes (1 ≤ A_i ≤ 1,000,000,000) in the first round and B_i votes (1 ≤ B_i ≤ 1,000,000,000) in the second round (if he or she makes it), determine which cow is expected to win the election. Happily for you, no vote count appears twice in the A_i list; likewise, no vote count appears twice in the B_i list.

Input

* Line 1: Two space-separated integers: N and K
* Lines 2..N+1: Line i+1 contains two space-separated integers: A_i and B_i

Output

* Line 1: The index of the cow that is expected to win the election.

Sample Input

Sample Output

#include<iostream>
#include<algorithm>
struct vote
{
int vo1;
int vo2;
int index;
};
int comp1(vote v1, vote v2)
{
return v1.vo1 > v2.vo1;
}
int comp2(vote v1, vote v2)
{
return v1.vo2 > v2.vo2;
}
using namespace std;
int main()
{
int n,k;
int i;
    vote in[50000];
    cin>>n;
    cin>>k;
for(i=0;i<n;i++)
    {
        cin>>in[i].vo1;
        cin>>in[i].vo2;
in[i].index=i+1;
    }
    sort(in,in+n,comp1);
    sort(in,in+k,comp2);
    cout<<in[0].index<<endl;
}

转载于:https://www.cnblogs.com/w0w0/archive/2011/11/23/2259864.html