uva 10012 How Big Is It?

Ian‘s going to California, and he has to pack his things, including his collection of circles. Given a set of circles, your program must find the smallest rectangular box in which they fit. All circles must touch the bottom of the box. The figure below shows an acceptable packing for a set of circles (although this may not be the optimal packing for these particular circles). Note that in an ideal packing, each circle should touch at least one other circle (but you probably figured that out).

Input

The first line of input contains a single positive decimal integer n, n<=50. This indicates the number of lines which follow. The subsequent n lines each contain a series of numbers separated by spaces. The first number on each of these lines is a positive integerm, m<=8, which indicates how many other numbers appear on that line. The nextm numbers on the line are the radii of the circles which must be packed in a single box. These numbers need not be integers.

Output

For each data line of input, excluding the first line of input containing n, your program must output the size of the smallest rectangle which can pack the circles. Each case should be output on a separate line by itself, with three places after the decimal point. Do not output leading zeroes unless the number is less than 1, e.g.0.543.

Sample Input

3
3 2.0 1.0 2.0
4 2.0 2.0 2.0 2.0
3 2.0 1.0 4.0

Sample Output

9.657
16.000
12.657

题目大意：给出一些圆的半径，求出可以装下这下这些圆的矩形箱子的最小尺寸（宽度）。圆都要与矩形底边相切。

解题思路：枚举所有情况，主要是将圆的半径转化为实际x坐标。

#include<stdio.h>  
#include<math.h>
#include<string.h>
#include<algorithm>
#include<stdlib.h>
using namespace std;  
double p[10], r[10], ans, temp;  
int n;
void getPos(int id)  
{  
	double x;  
	for (int i = 0; i < id; ++i)  
	{  
		x = sqrt((r[i] + r[id]) * (r[i] + r[id]) - (r[i] - r[id]) * (r[i] - r[id])) + p[i];  
		p[id] = max(p[id], x);  
	}  
}  
int main()  
{  
	int T;  
	scanf("%d", &T);  
	while (T--)  
	{  
		scanf("%d", &n);  
		for (int i = 0; i < n; i++) scanf("%lf", &r[i]);  
		sort(r, r + n);  
		ans = 0xFFFFFFFF;  
		do
		{  
			memcpy(p, r, sizeof(p));  
			for (int i = 1; i < n; i++) {
				getPos(i); //计算该圆圆心的x轴坐标  
			}
			temp = 0.0;  
			for (int i = 0; i < n; i++) {
				temp = max(temp, r[i] + p[i]);   
			}
			ans = min(ans, temp);  
		}  
		while (next_permutation(r, r + n)); //枚举不同排列顺序 
		printf("%.3f\n", ans);  
	}  
	return 0;  
}

uva 10012 How Big Is It?（枚举）

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