SGU114-Telecasting station

114. Telecasting station

time limit per test: 0.5 sec.
memory limit per test: 4096 KB

 

Every city in Berland is situated on Ox axis. The government of the country decided to build new telecasting station. After many experiments Berland scientists came to a conclusion that in any city citizens displeasure is equal to product of citizens amount in it by distance between city and TV-station. Find such point on Ox axis for station so that sum of  displeasures of all cities is minimal.

 

Input

Input begins from line with integer positive number N (0<N<15000) – amount of cities in Berland. Following N pairs (X, P) describes cities (0<X, P<50000), where X is a coordinate of city and P is an amount of citizens. All numbers separated by whitespace(s).

 

Output

Write the best position for TV-station with accuracy 10^-5.

 

Sample Input

4
1 3
2 1
5 2
6 2

 

Sample Output

3.00000

 

 

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题意是说，有个国家的城市之间要建个公交（我英语不好、翻译可能会跑偏、也可能是电视塔、也可能是别的、我也不知道是啥），然后有一个不满意度，咱们自己想想也知道，肯定都喜欢把地铁站设在自己家门口。然后这个不满意度=城市的人数*这个站的位置。。。然后的然后、我也不会做这个题，直接上理论：数学家说，这是一个带权中位数问题，具体内容见 http://baike.baidu.com/link?url=NZbFTsJCCn_jsx8W24aLu6XGEVdpDH0hpO_SIzeCF7vFi_Nz-xV4pe7tzuJBpQsQSzoa719FmHYb3lr7QKV3q_   这个讲的很明白了。然后，下面的代码就自然手到擒来了！！！

 

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
struct N
{
    double x;
    int p;
} a[15009];
int b[15009];
int cmp(N a,N b)
{
    return a.x<b.x;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(b,0,sizeof(b));
        for(int i=0; i<n; i++)
            scanf("%d%d",&a[i].x,&a[i].p);
        sort(a,a+n,cmp);
        int s=0;
        for(int i=0; i<n; i++)
        {
            s+=a[i].p;
            b[i]=s;
        }
        int mid=s/2,w;
        for(int i=0; i<n; i++)
        {
            if(b[i]>=mid)
            {
                w=i;
                break;
            }
        }
        printf("%d\n",a[w].x);
    }
    return 0;
}