Party All the Time

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4282    Accepted Submission(s): 1355

Problem Description

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.

Input

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 : X_i,W_i, representing the location and the weight of the i-th spirit. ( |x_i|<=10⁶, 0<w_i<15 )

Output

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.

Sample Input

1

4

0.6 5

3.9 10

5.1 7

8.4 10

Sample Output

Case #1: 832

Author

[email protected]_Goldfinger

Source

2012 Multi-University Training Contest 6

题目大意：n个人要在某一条线段某位置聚会，这n个人开始有一个坐标xi，一个体重wi，每个人到该位置距离设为S。让你算所有人到该位置的S^3*w的和最小。求出最小和。

解题思路：套用三分求解。

#include<bits/stdc++.h>
using namespace std;
typedef long long INT;
const int maxn=1e5+200;
int n;
const int Mv=1e6;
const double eps=1e-2;
struct Spirit{
    double x;
    double w;
}spirits[5*maxn];
double Abs(double xx){
    return xx>0?xx:-xx;
}
double Pow(double x,int nn){
    double ret=1.0;
    for(int i=1;i<=nn;i++)
        ret*=x;
    return ret;
}
double cal(double xx){
    double sum=0;
    for(int i=1;i<=n;i++){
        sum+=Pow(Abs(spirits[i].x-xx),3)*spirits[i].w;
    }
    return sum;
}
double three_div(double L,double R){    //三分求最值
    double mid=(L+R)/2,mid_L=(L+mid)/2;
    while(Abs(cal(mid)-cal(mid_L))>eps){ //条件应该视情况而定
        mid=(R+L)/2.0;
        mid_L=(L+mid)/2.0;
        if(cal(mid)>cal(mid_L)){
            R=mid;
        }else{
            L=mid_L;
        }
    }
    return mid; //得到最值的坐标位置
}
int main(){
    int t,cnt=0;
    scanf("%d",&t);
    double min_v=Mv*(-1.0),max_v=Mv*1.0;
    while(t--){
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%lf%lf",&spirits[i].x,&spirits[i].w);
        }
        double xx=three_div(min_v,max_v);
        printf("Case #%d: %lld\n",++cnt,(INT)(cal(xx)+0.5));
    }
    return 0;
}