poj——3664——Election Time

poj——3664——Election Time"/>

poj——3664——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

#include<iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int n,k;
const int M=50010;
struct node
{int a;int b;int num;
}Node[M];
int cmpa(node a,node b)
{if(a.a!=b.a)return a.a>b.a;elsereturn a.b>b.b;
}
int cmpb(node a,node b)
{if(a.b!=b.b)return a.b>b.b;elsereturn a.a>b.a;
}
int main()
{while(cin>>n>>k){for(int i=0;i<n;i++){scanf("%d%d", &Node[i].a, &Node[i].b);Node[i].num=i+1;}sort(Node,Node+n,cmpa);sort(Node,Node+k,cmpb);printf("%d\n",Node[0].num);}return 0;
}