POJ 3122 分披萨（二分查找）_poj3122

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;
        else
            r_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;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50

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;
        else
            s_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;
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60