三分求极值

1142 : 三分·三分求极值

题目传送：hihoCoder - 1142 - 三分·三分求极值

二分适用于单调函数，对于需要逼近的区间做二等分，来求解某点的值等。
三分适用于凸形函数，对于需要逼近的区间做三等分。

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;double a, b, c, x, y;double get_dist(double xx) {    double yy = a * xx * xx + b * xx + c;    return sqrt((xx - x) * (xx - x) + (yy - y) * (yy - y));}int main() {    double l = -500, r = 500;    scanf("%lf %lf %lf %lf %lf", &a, &b, &c, &x, &y);    while(r - l > 0.000001) {//这里为了保证精度，尽量开小点，因为是分治，即使再小也能够承受        double d = (r - l) / 3;        double rm = r - d;        double lm = l + d;        if(get_dist(lm) < get_dist(rm)) {            r = rm;        }        else l = lm;    }    printf("%.3lf\n", get_dist(l));    return 0;}