Election Time

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 Ai votes (1 ≤ Ai ≤ 1,000,000,000) in the first round and Bi votes (1 ≤ Bi ≤ 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 Ai list; likewise, no vote count appears twice in the Bi list.

Input

• Line 1: Two space-separated integers: N and K
• Lines 2..N+1: Line i+1 contains two space-separated integers: Ai and Bi

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
HINT

题意：有n头牛，第一列表示第一轮Ai的投票，第二列表示第二轮Bi的投票，在第一轮排名前k的可以进入第二轮，然后在第二轮找最大的票数，输出牛的序号即可

第一列从大到小排序，在第二列前k个里找个最大的。over

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<stdlib.h>
#include<algorithm>
#define sf(a)   scanf("%d",&a)
#define sff(a,b)   scanf("%d %d",&a,&b)
#define pf(ans)   printf("%d\n",ans)
using namespace std;
const int M=50000+10;
struct node
{int Ai;int Bi;int sub;
} nod[M+50];
bool cmp(struct node p,struct node q)
{return p.Ai>q.Ai;
}
int main()
{int n,k;sff(n,k);int i;for(i=1; i<=n; i++){sff(nod[i].Ai,nod[i].Bi);nod[i].sub=i;}sort(nod+1,nod+n+1,cmp);int Max=0,sub=0;for(i=1; i<=k; i++){if(nod[i].Bi>Max){Max=nod[i].Bi;sub=nod[i].sub;}}pf(sub);
}