HDU 4355 Party All the Time (三分求凸函数极值)

题目大意：有n个精灵位于一维坐标轴上，每个精灵有一个权值w，每个精灵走到另一个位置耗费能量S^3 * w。（s是两点间距离）。现在精灵要聚会，求选取一点，使所有精灵到这点浪费能量之和最小。分析：设这点的位置是x，精灵位置是pi，则浪费能量值和f = sigma[(pi-x)^3*wi].其中x是变量. 通过对这个函数求二次导可以发现，这个函数是一个下凸函数.可以用三分做. 具体的三分方法如图：凸函数：

凹函数：

三分求凸函数极值模板：

//如果一个解函数的二次导恒>0(or <0)，则该函数为凸函数.

//精度模板
const double eps = 1e-6;
bool dy(double x,double y)  {   return x > y + eps;} 		// x > y
bool xy(double x,double y)  {   return x < y - eps;} 		// x < y
bool dyd(double x,double y) {   return x > y - eps;} 		// x >= y
bool xyd(double x,double y) {   return x < y + eps;}     	// x <= y
bool dd(double x,double y)  {   return fabs( x - y ) < eps;}    // x == y

double cal(double pos){
    /* 根据题目的意思计算 */
}

double solve(double left, double right){
    while(xy(left, right)){
        double mid = DMID(left, right);
        double midmid = DMID(mid, right);
        double mid_area = cal(mid);
        double midmid_area = cal(midmid);
        if (xyd(mid_area, midmid_area))     //下凸函数，上凸函数用dyd(mid_area, midmid_area)
            right = midmid;
        else
            left = mid;
    }
    return left;
}

HDU 4355 代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ((x+y)>>1)
#define DMID(x,y) ((x+y)/2)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

const int N = 50002;
double x[N], w[N];
int n;

//二分精度模板
const double eps = 1e-6;
bool dy(double x,double y)  {   return x > y + eps;} 		// x > y
bool xy(double x,double y)  {   return x < y - eps;} 		// x < y
bool dyd(double x,double y) {   return x > y - eps;} 		// x >= y
bool xyd(double x,double y) {   return x < y + eps;}     	// x <= y
bool dd(double x,double y)  {   return fabs( x - y ) < eps;}    // x == y

double cal(double pos){
    /* 根据题目的意思计算 */
    double res = 0.0;
    for (int i = 0; i < n; i ++){
        res += abs(pos - x[i]) * abs(pos - x[i]) * abs(pos - x[i]) * w[i];
    }
    return res;
}

double solve(double left, double right){
    while(xy(left, right)){
        double mid = DMID(left, right);
        double midmid = DMID(mid, right);
        double mid_area = cal(mid);
        double midmid_area = cal(midmid);
        if (xyd(mid_area, midmid_area))     //下凸函数，上凸函数dyd(mid_area, midmid_area)
            right = midmid;
        else
            left = mid;
    }
    return left;
}

int main(){
    int t;
    scanf("%d", &t);
    for (int caseo = 1; caseo <= t; caseo ++){
        scanf("%d", &n);
        double low = -1000000, high = 1000000;
        for (int i = 0; i < n; i ++){
            scanf("%lf %lf", &x[i], &w[i]);
            low = min(low, x[i]);
            high = max(high, x[i]);
        }

        printf("Case #%d: %.0f\n", caseo, (cal(solve(low, high))));
    }
	return 0;
}

举杯独醉，饮罢飞雪，茫然又一年岁。 ------AbandonZHANG

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原文地址：https://www.cnblogs.com/AbandonZHANG/p/4114015.html