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You are given n cubes, each cube is described by two points in 3D space: (x1, y1, z1) being one corner of the cube and (x2, y2, z2) being the opposite corner. Assume that the sides of each of the cubes are parallel to the axis. Your task is to find the volume of their intersection.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 100). Each of the next n lines contains six integers x1 y1 z1 x2 y2 z2 (1 ≤ x1, y1, z1, x2, y2, z2 ≤ 1000, x1 < x2, y1 < y2, z1 < z2) where (x1, y1, z1) is the co-ordinate of one corner and (x2, y2, z2) is the co-ordinate of the opposite corner.

Output

For each case, print the case number and volume of their intersection.

Sample Input

2

2 1 1 1 3 3 3 1 1 1 2 2 2 3 7 8 9 20 20 30 2 2 2 50 50 50 13 14 15 18 30 40 Output for Sample Input

Case 1: 1

Case 2: 450

给出底下的坐标和与其相对的坐标，求立方体的交

交集的底坐标由最大的那个决定，顶坐标反之

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            #define INF 0x3f3f3f3f#define MAXN 2005#define Mod 10001using namespace std;int up[MAXN],down[MAXN],tup[MAXN],tdown[MAXN];int main(){ int t,cnt=1; scanf("%d",&t); while(t--) { int n; scanf("%d",&n); for(int i=0; i<3; ++i) scanf("%d",&down[i]); for(int i=0; i<3; ++i) scanf("%d",&up[i]); for(int j=0; j
            
             up[i]) flag=true; area*=(up[i]-down[i]); } if(!flag) printf("%d\n",area); else printf("0\n"); } return 0;}
            
           
          
         
        
       
      
     
    
   

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