三分法：Party all the time

Party All the Time

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 829 Accepted Submission(s): 308



[align=left]Problem Description[/align]
In the Dark forest, there is a Fairy kingdom where all the spirits will go together and Celebrate the harvest every year. But there is one thing you may not know that they hate walking so much that they would prefer to stay at home
if they need to walk a long way.According to our observation,a spirit weighing W will increase its unhappyness for S3*W units if it walks a distance of S kilometers.

Now give you every spirit's weight and location,find the best place to celebrate the harvest which make the sum of unhappyness of every spirit the least.

[align=left]Input[/align]
The first line of the input is the number T(T<=20), which is the number of cases followed. The first line of each case consists of one integer N(1<=N<=50000), indicating the number of spirits. Then comes N lines in the order that
x[i]<=x[i+1] for all i(1<=i<N). The i-th line contains two real number : Xi,Wi, representing the location and the weight of the i-th spirit. ( |xi|<=106, 0<wi<15 )

[align=left]Output[/align]
For each test case, please output a line which is "Case #X: Y", X means the number of the test case and Y means the minimum sum of unhappyness which is rounded to the nearest integer.

[align=left]Sample Input[/align]

1
4
0.6 5
3.9 10
5.1 7
8.4 10

[align=left]Sample Output[/align]

Case #1: 832

凸函数，直接用三分法求解

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

#define EPS 1e-7
#define MAXN 50050
double x[MAXN], w[MAXN];
int n;

double Cal(double pos){
double ans = 0;
for(int i = 1; i <= n; i ++){
double tmp = abs(pos - x[i]);
ans += w[i] * tmp * tmp * tmp;
}
return ans;
}

double Solve(){
double Left, Right;
double mid, midmid;
double mid_value, midmid_value;
Left = x[1];
Right = x
;
while(Left + EPS < Right){
mid = (Left + Right) / 2;
midmid = (mid + Right) / 2;
mid_value = Cal(mid);
midmid_value = Cal(midmid);
if(mid_value <= midmid_value)
Right = midmid;
else
Left = mid;
}
return Cal(Left);
}

int main(){
int t;
int cnt = 0;
scanf("%d", &t);
while(t --){
cnt ++;
scanf("%d", &n);
for(int i = 1; i <= n; i ++){
scanf("%lf %lf", &x[i], &w[i]);
}
printf("Case #%d: %.0lf\n", cnt, Solve());
}
return 0;
}