SGU 114 Telecasting station

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Description

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

用三分来做是可以的。

因为函数f一定是凹函数。

代码：

#include<iostream>
#include<string.h>
using namespace std;

int list1[15001];
int list2[15001];
int n;

double f(double p)
{
double sum = 0;
for (int i = 0; i < n; i++)

sum += list2[i] * ((list1[i]>p) ? list1[i] - p : p - list1[i]);
return sum;
}

int main()
{
cin >> n;
for (int i = 0; i < n; i++)cin >> list1[i] >> list2[i];
double low = 0, high = 50000;
double mid, mm;
while (low + 0.000001 < high)
{
mid = (low + high) / 2;
mm = (mid + high) / 2;
if (f(mid) < f(mm))high = mm;
else low = mid;
}
cout << mid;
return 0;
}

实际上，这个问题要么有唯一解，要么有无穷解。

本题就是要找一个点，使得左边的人和右边的人一样多。

这个点可能是所给的某个点，也有可能是某2个相邻的点之间的线段上 任意一点都行。

也就是说，一定可以取所给的某个点作为答案。

但是这没什么卵用啊，那么多点，如果不排序的话就只能把每个点试一下，不管是哪种都不会很快。