对结构体的快速排序问题（用库函数）

题目链接：sdut 2695

Election Time

Time Limit: 1000MS Memory limit: 65536K

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

两次快排就行，用c++中的库函数，很快就能做出来，关键就是如何对结构体排序的问题：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
struct vode
{
    int u,v,flag;
}f[500000];
int cmp(const void *a,const void *b)
{
    return ((struct vode *)b)->u-((struct vode *)a)->u;
}
int cp(const void *a,const void *b)
{
    return ((struct vode *)b)->v-((struct vode *)a)->v;
}
int main()
{
    int i;
    for(i=0;i<=50000;i++)
    {
        f[i].u=0;
        f[i].v=0;
        f[i].flag=i;
    }
    int m,n;
    scanf("%d%d",&m,&n);
    for(int i=0;i<=m-1;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        f[i].u=u;
        f[i].v=v;
    }
    qsort(f,m,sizeof(f[0]),cmp);
    qsort(f,n,sizeof(f[0]),cp);
    printf("%d",f[0].flag+1);
}