HDU 4355.Party All the Time-三分-优快云博客

本文介绍了一个有趣的问题：在一个精灵王国中，如何找到一个使所有精灵因行走距离导致的不快乐总和最小的聚会地点。文章提供了一种通过三分法搜索最佳位置的方法，并给出了完整的C++代码实现。

Party All the Time

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5462 Accepted Submission(s): 1707

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 S ³*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|<=10 ⁶, 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

0.6 5

3.9 10

5.1 7

8.4 10

Sample Output

Case #1: 832

直接代码了

#include<bits/stdc++.h>
using namespace std;
#define N 50500
struct node{
    double p;
    double w;
}a[N];
int n;
double sum(double mid){
    double sum=0.0;
    for(int i=0;i<n;i++){
        double s=fabs(a[i].p-mid);
        sum+=s*s*s*a[i].w;
    }
    return sum;
}
double sanfen(double l,double r){
    double mid,midd,ans1,ans2,left,right;
    left=l;right=r;
    while(left+0.000000001<right){
        mid=(left+right)/2.0;
        midd=(mid+right)/2.0;
        ans1=sum(mid);
        ans2=sum(midd);
        if(ans1<ans2)right=midd;
        else left=mid;
    }
    return left;
}
int main(){
    int t,num;
    double left,right,ans,m;
    num=0;
    while(~scanf("%d",&t)){
      while(t--){
            num++;
            left=1e5;right=1e-5;
            scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&a[i].p,&a[i].w);
            left=min(left,a[i].p);
            right=max(right,a[i].p);
        }
        m=sanfen(left,right);
        ans=sum(m);
        printf("Case #%d: %.0lf\n",num,ans);
     }
    }
     return 0;
}