SDUT 2886 Weighted Median

Weighted Median

                    Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Description

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?

 

Input

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).

 

Output

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

 

Sample Input

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

Sample Output

20

Hint

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

给出一个数列，对于数列中的每一个数都有对应的权值，将所有的权值加起来除以2.求以哪个数为界可满足题目要求。

解题思路：先按从小到大的顺序排序，当Xk为第一个数时，小于Xk的和最小，大于Xk的和最大。当Xk逐渐向右移时，小于Xk的和逐渐变大，大于Xk的和逐渐变小。也就是说，越向右移动，越能满足     这个条件。再看这一侧，虽然变得越来越大，会面临突破S/2的一刻，但突破之前一直是满足题目条件的。

所以说，在突破的那一刻，该数的右侧所有数的和，为尽可能的小，该数左侧所有数的和，满足题目所给条件，即该点为Xk。

某一个数将一个数列分成两块，随着前面的变化，后面也会有规律的变化。

AC代码：

#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct number{
    int number;
    int weight;
}stu[10000009];
bool cmp(struct number a,struct number b)
{
    if(a.number==b.number)
        return a.weight<b.weight;
    return a.number<b.number;
}
int main()
{
    int i,t;
    long long int y,ab;
    double mid;
    while(scanf("%d",&t)!=EOF)
    {
        y=0;
        ab=0;
        for(i=0;i<=t-1;i++)
            scanf("%d",&stu[i].number);
        for(i=0;i<=t-1;i++)
        {
            scanf("%d",&stu[i].weight);
            y+=stu[i].weight;
        }
        mid=y/2.0;
        sort(stu,stu+t,cmp);
        for(i=0;i<=t-1;i++)
        {
            ab+=stu[i].weight;
            if(ab>=mid)
            {
                printf("%d\n",stu[i].number);
                break;
            }
        }

    }
    return 0;
}