[ACM] SDUT 2886 Weighted Median (省赛最悲剧的一道题）

Weighted MedianTime 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 satisfyingand, 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

710 35 5 10 15 5 2010 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

解题思路：

当时最后半个多小时三个人死活没想出来怎么做，，现在拿出这道题一看十几分钟就解决了。。。现场的时候心态实在是太重要了，一慌脑子就容易空白。。。

代码：

#include <iostream>#include <algorithm>#include <stdio.h>using namespace std;const int maxn=1e7+2;int n;struct Node{int x,w;}node[maxn];bool cmp(Node a,Node b){if(a.x<b.x)return true;return false;}int main(){while(scanf("%d",&n)!=EOF){long long sum=0;for(int i=1;i<=n;i++)scanf("%d",&node[i].x);for(int i=1;i<=n;i++){scanf("%d",&node[i].w);sum+=node[i].w;}long double S=sum*0.5;sort(node+1,node+1+n,cmp);long long xiao=0,da=0;int ans;for(int i=1;i<=n-1;i++){xiao+=node[i].w;da=sum-xiao-node[i+1].w;if(xiao<S&&da<=S){ans=node[i+1].x;break;}}cout<<ans<<endl;}return 0;}

有时我们选择改变，并非经过深思熟虑，