POJ 3122 分披萨（二分查找）

题目链接：http://poj.org/problem?id=3122

题目大意：

有 n 块披萨（大小不一样）， f 个人分，包含主人自己 f+1 人；
每人吃的披萨必须是一块披萨上切下来的。每个人吃的披萨相等，披萨可以有剩余。求每人吃的最大披萨面积。

思路：

假设每人分得的披萨面积等效为半径 R的圆；
每块披萨可以分给几个人呢？ r[i] 表示披萨半径，则是 r[i]²/R² 取整个人
然后全部累加起来，如果总和大于等于 f+1，则每个人还有分更大的披萨的可能，R取值增大
如果总和小于 f+1，则每个人分的太大了，不够分的，R取值减小
R的取值范围在（0，max（r[i]））

Wrong Answer代码

/*** @description: 有 n 块披萨（大小不一样）， f 个人分，包含主人自己 f+1 人；*                  每人吃的披萨必须是一块披萨上切下来的。求* @author: michael ming* @date: 2019/4/20 0:23* @modified by: */
#include <iostream>
#include <iomanip>
#define PI 3.14159265359
using namespace std;
const double error = 1e-7;
double find_max_R(size_t pizza_num, int *r_pizza, double r_low, double r_high, size_t people)
{double R_we_want = r_low+(r_high-r_low)/2;size_t people_get_pizza = 0;while(r_high - r_low > error){people_get_pizza = 0;R_we_want = r_low+(r_high-r_low)/2;for(int i = 0; i < pizza_num; ++i)people_get_pizza += (int)(r_pizza[i]*r_pizza[i]/(R_we_want*R_we_want));if(people_get_pizza >= people)r_low = R_we_want;elser_high = R_we_want;}return R_we_want;
}
int main()
{size_t t, pizza_num, friend_num;double r_max_pizza = 0;cin >> t;while(t--){cin >> pizza_num >> friend_num;int *r_pizza = new int [pizza_num];for(int i = 0; i < pizza_num; ++i){cin >> r_pizza[i];r_max_pizza = r_pizza[i] > r_max_pizza ? r_pizza[i] : r_max_pizza;}r_max_pizza = find_max_R(pizza_num,r_pizza,0,r_max_pizza,friend_num+1);cout << setiosflags(ios::fixed) << setprecision(4) << PI*r_max_pizza*r_max_pizza << endl;delete[] r_pizza;r_pizza = NULL;}return 0;
}

在这里插入图片描述

AC代码（主要修改，精度问题，把求人数的除法改成减法）

/*** @description: 有 n 块披萨（大小不一样）， f 个人分，包含主人自己 f+1 人；*                  每人吃的披萨必须是一块披萨上切下来的。求* @author: michael ming* @date: 2019/4/20 0:23* @modified by: */
#include <iostream>
#include <iomanip>
#include <math.h>
#include <algorithm>
#define PI acos(-1.0)
using namespace std;
const double error = 1e-7;
double find_max_R(size_t pizza_num, double *s_pizza, double s_low, double s_high, size_t people)
{double S_we_want = s_low+(s_high-s_low)/2.0;size_t people_get_pizza = 0;while(s_high - s_low > error){people_get_pizza = 0;S_we_want = s_low+(s_high-s_low)/2.0;for(int i = 0; i < pizza_num; ++i){double temp = s_pizza[i];while(temp-S_we_want>=0){temp -= S_we_want;	//改成减法，不易丢失精度people_get_pizza++;}}if(people_get_pizza >= people)s_low = S_we_want;elses_high = S_we_want;}return S_we_want;
}
int main()
{size_t t, pizza_num, friend_num;double s_max_pizza = 0.0;cin >> t;while(t--){cin >> pizza_num >> friend_num;double *s_pizza = new double [pizza_num];for(int i = 0; i < pizza_num; ++i){cin >> s_pizza[i];s_pizza[i] *= s_pizza[i];}sort(s_pizza, s_pizza+pizza_num);s_max_pizza = find_max_R(pizza_num,s_pizza,0,s_pizza[pizza_num-1],friend_num+1);cout << setiosflags(ios::fixed) << setprecision(4) << PI*s_max_pizza << endl;delete[] s_pizza;s_pizza = NULL;}return 0;
}