POJ 3664 Election Time

Election Time

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

5 3
3 10
9 2
5 6
8 4
6 5

Sample Output

5

水题，结构体排序，要按第一次得票和第二次得票排两次序，第一次对n个排序，第二次对前k个排序。

#include <stdio.h>
#include <algorithm>
using namespace std;
typedef __int64 ll;

const int maxn=50000+10;

struct pp
{
	ll first,second;
	int id;
}s[maxn];

int cmp1(const pp &x,const pp &y){
	return x.first>y.first;
}

int cmp2(const pp &x,const pp &y){
	return x.second>y.second;
}

int main()
{
	int n,k,i,j,ans;
	scanf("%d%d",&n,&k);
	for(i=0;i<n;i++){
		scanf("%I64d%I64d",&s[i].first,&s[i].second);
		s[i].id=i+1;
	}
	sort(s,s+n,cmp1);
	sort(s,s+k,cmp2);		//注意第二次是前k个
	printf("%d\n",s[0].id);
	return 0;
}