2014山东省第五届ACM省赛 Weighted Median

Weighted Median

Time Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

For n elements x1, x2, ..., xn with positive integer weights w1, w2, ..., wn. The weighted median is the element xk satisfying
 and   , S indicates 

Can you compute the weighted median in O(n) worst-case?

 

输入

There are several test cases. For each case, the first line contains one integer n(1 ≤  n ≤ 10^7) — the number of elements in the sequence. The following line contains n integer numbers xi (0 ≤ xi ≤ 10^9). The last line contains n integer numbers wi (0 < wi < 10^9).

 

输出

One line for each case, print a single integer number— the weighted median of the sequence.

 

示例输入

7
10 35 5 10 15 5 20
10 35 5 10 15 5 20

示例输出

20

提示

The S which indicates the sum of all weights may be exceed a 32-bit integer. If S is 5,  equals 2.5.

来源

2014年山东省第五届ACM大学生程序设计竞赛

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

struct ss
{
    long long int x;
    long long int w;
};

int cmp(ss a,ss b)
{
    return a.x<b.x;
}

const int maxn = 10000000+1;
ss a[maxn];
int main()
{
    int  n;

    while(cin>>n)
    {
        long long sum=0,w;
        for(int i=0; i<n; i++)
        {
            a[i].x=0;
            a[i].w=0;
        }
        for(int i=0; i<n; i++)
        {
            cin>>a[i].x;
        }
        for(int i=0; i<n; i++)
        {
            cin>>w;
            a[i].w=w*2;
            sum+=w*2;
        }
        sum/=2;
        sort(a,a+n,cmp);
        long long int s=0;
        long long int q=a[0].x;
        for(int i=0; i<n; i++)
        {
            s+=a[i].w;
            if(s>=sum)
            {
                q=a[i].x;
                break;
            }

        }
        cout<<q<<endl;
    }
    return 0;
}