HDU 4355 Party All the Time（三分）

参考链接：
https://blog.csdn.net/Littlewhite520/article/details/70144763
https://blog.csdn.net/u013557725/article/details/38377289
三分简述
注意区分二分和三分，如果求单调函数描述的问题的最优解，可以用二分；但是如果是个不单调函数，就需要用到三分。三分用于凸性函数（凹凸）的最优解问题，说白了就是找那个凸点所在的位置或者它所附加的信息。（二分和三分的模板性都比较强，都是一个while中不断缩小l和r的距离和一个判断某个点是否满足条件的函数组成）
下面这个图就是描述怎么缩小l和r之间的距离

查找方法：以先增后减型为例（找了很多博客，int类型和double类型的查找不太一样）

int sear(int l, int r)
{while (l<r-1){int mid1=(l+r)/2;int mid2=(mid1+r)/2;if (check(mid1)>check(mid2))r=mid2;elsel=mid1;}return check(l)>check(r)? l: r;
}

double sear(double l, double r)
{while (fabs(r-l)>=eps){//大于 还是大于等于都行double mid1=l+fabs(r-l)/3;double mid2=l+fabs(r-l)/3*2;if (check(mid1)>check(mid2))r=mid2;elsel=mid1;}return l;//因为误差很小了，所以返回哪个变量都一样
}

HDU 4355 Party All the Time
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
1
4
0.6 5
3.9 10
5.1 7
8.4 10
Sample Output
Case #1: 832

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;const double eps=1e-5;
double p[50100], w[50100];
int n;
double cal(double x)
{double ans=0;for (int i=1; i<=n; i++){double t=fabs(p[i]-x);ans+=t*t*t*w[i];}return ans;
}
double sear(double l, double r)
{double mid1, mid2;while (fabs(r-l)>eps){mid1=l+(fabs(r-l))/3;mid2=l+(fabs(r-l))/3*2;if (cal(mid1)>cal(mid2))l=mid1;elser=mid2;}return l;
}
int main()
{int T, cas=1;scanf("%d", &T);while (T--){scanf("%d", &n);double l=999999, r=-999999;for (int i=1; i<=n; i++){scanf("%lf%lf", &p[i], &w[i]);l=min(l, p[i]);r=max(r, p[i]);}double point=sear(l, r);double ans=cal(point);printf("Case #%d: %.0f\n", cas++, ans);}return 0;
}

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