SGU 114 Telecasting station（二分）

114. Telecasting station

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 citizensdispleasure 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

41 32 15 26 2

Sample Output

3.00000

题目大意：有一些城市坐落在x轴上，每个城市i都有一个人口数Pi。现在要在x轴上建一座电视塔，城市居民的烦恼值等于城市与电视塔的距离和城市人口的乘积，求一个电视塔的位置，使得所有城市的总烦恼值最小。

解题思路：三分法求极小值。（样例输出是3，自己写的程序跑出来是5，还以为自己错了，调试了半天。。。）

代码如下：

#include <algorithm>#include <cctype>#include <climits>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <queue>#include <set>#include <stack>#include <vector>#define EPS 1e-6#define INF INT_MAX / 10#define LL long long#define MOD 100000000#define PI acos(-1.0)const int maxn = 15005;struct city{    int x;    int p;};city ct[maxn];int n;double cal(double dis){    double res = 0;    for(int i = 1;i <= n;i++){        res += fabs((double)ct[i].x - dis) * (double)ct[i].p;    }    return res;}int main(){    scanf("%d",&n);    double lb,lm,um,ub = 0;    for(int i = 1;i <= n;i++){        scanf("%d %d",&ct[i].x,&ct[i].p);        if(ct[i].x > ub)            ub = ct[i].x;    }    lb = 0;    while(ub - lb > EPS){        lm = (lb + ub) / 2.0;        um = (lm + ub) / 2.0;        if(cal(lm) < cal(um))            ub = um;        else            lb = lm;    }    printf("%.5f\n",lb);}