题目意思： 给定m个圆的半径，现在要求找到一个矩形使得每一个球都以地面相切，要求输出最小的矩阵的长度
1 #include <iostream> 2 #include <algorithm> 3 #include <cstring> 4 #include <cstdlib> 5 #include <cstdio> 6 #include <cmath> 7 using namespace std; 8 int num; 9 bool vis[10]; 10 double a[8], c[8], l[8], _max; 11 double sqr(double r) 12 { 13 return r*r; 14 } 15 16 void dfs(int cnt, double len) 17 { 18 if(len > _max) 19 return ; 20 if(cnt == num) 21 { 22 double ll = c[0]; 23 len += c[cnt - 1]; 24 for(int i = 0; i < num - 1; i++) 25 if(l[i] + c[i] > len) 26 len = l[i] + c[i]; 27 for(int i = 1; i < num; i++) 28 if(c[i] - l[i] > ll) 29 ll = c[i] - l[i]; 30 len += ll; 31 if(len < _max) 32 _max = len; 33 return ; 34 } 35 for(int i = 0; i < num; i++) 36 { 37 if(vis[i]) continue; 38 vis[i] = true; 39 c[cnt] = a[i]; 40 l[cnt] = c[cnt]; 41 for(int j = cnt - 1; j >= 0; j--) 42 { 43 double dis = l[j] + sqrt(sqr(c[j]+c[cnt])-sqr(c[j]-c[cnt])); 44 if(dis > l[cnt]) 45 l[cnt] = dis; 46 } 47 dfs(cnt + 1, l[cnt]); 48 vis[i] = false; 49 } 50 } 51 int main() 52 { 53 int n; 54 scanf("%d", &n); 55 memset(vis, 0, sizeof(vis)); 56 while(n--) 57 { 58 _max = 0x7FFFFFFF; 59 scanf("%d", &num); 60 for(int i = 0; i < num; i++) 61 scanf("%lf", &a[i]); 62 for(int i = 0; i < num; i++) 63 { 64 vis[i] = true; 65 l[0] = 0; 66 c[0] = a[i]; 67 dfs(1, 0); 68 vis[i] = false; 69 } 70 printf("%.3lf ", _max); 71 } 72 return 0; 73 }