poj 3664 Election Time

题目描述

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.

输入

 Line 1: Two space-separated integers: N and K 

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

 

输出

 The index of the cow that is expected to win the election.

示例输入

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

示例输出

5

题意：输入N行数据 求出前k项中第二轮得票最多的是哪一项

#include<stdio.h>
#include<stdlib.h>
#define N 50005

typedef struct cow
{
    int ai,bi,data;//ai第一列数据，bi第二列数据，data存储位置信息
} Cow;
Cow a[N];

/*int型数据qsort排序方法头文件位于#include<stdlib.h>
格式:
int cmp ( const void *a , const void *b )
{
return *(int *)a - *(int *)b;由小到大排序
}
qsort(num,100,sizeof(num[0]),cmp);*/

int cmp ( const void *a , const void *b )
{
    return *(int *)b - *(int *)a;
}

int main()
{
    int n,k,i,max,j;
    scanf("%d%d",&n,&k);
    for(i=0; i<n; i++)
    {
        scanf("%d%d",&a[i].ai,&a[i].bi);
        a[i].data=i+1;
    }
    qsort(a,n,sizeof(Cow),cmp);//int型数据qsort排序方法头文件位于#include<stdlib.h>
    for(i=0,max=0; i<k; i++)
    {
        if(max<a[i].bi)
        {
            max = a[i].bi;
            j = i;
        }
    }
    printf("%d
",a[j].data);
    return 0;

}