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).
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 integer m, m<=8,
which indicates how many other numbers appear on that line. The next m numbers on the line are the radii of the circles which must be packed in a single box. These numbers need not be
integers.
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.
3
3 2.0 1.0 2.0
4 2.0 2.0 2.0 2.0
3 2.0 1.0 4.0
9.657
16.000
12.657
題意：
給出幾個圓的半徑，貼着底下排放在一個長方形裏面，求出如何擺放能使長方形底下長度最短;
#include<cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
double a[10], r[10];
///計算圓心的實際橫座標
inline void calc(int id)
{
double x;
for (int i = 0; i < id; ++i)
{
x = sqrt((a[i] + a[id]) * (a[i] + a[id]) - (a[i] - a[id]) * (a[i] - a[id])) + r[i];
r[id] = max(r[id], x);
}
}
int main()
{
int T, n, i;
double ans, tmp;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
for (i = 0; i < n; ++i) scanf("%lf", &a[i]);
sort(a, a + n);
ans = 1e100;
do
{
memcpy(r, a, sizeof(a));
for (i = 1; i < n; ++i) calc(i);///計算圓心的實際橫座標
tmp = 0.0;
for (i = 0; i < n; ++i) tmp = max(tmp, r[i] + a[i]);///這麼算是因爲不一定最右邊那個就靠在箱子右側
ans = min(ans, tmp);
}
while (next_permutation(a, a + n));
printf("%.3f\n"
, ans);
}
return 0;
}
</pre><pre name="code" class="cpp"><pre class="sh-cpp sh-sourceCode" style="white-space: pre-wrap; word-wrap: break-word; margin-top: 0px; margin-bottom: 0px; padding: 5px 5px 5px 7px; overflow: auto; font-size: 16px; line-height: 17.6000003814697px; font-family: 'Courier New', Courier, monospace;">
