1188 Gleaming the Cubes

Gleaming the Cubes

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 428		Accepted: 213

Description
As chief engineer of the Starship Interprize, the task of repairing the hyperstellar, cubic, transwarped-out software has fallen on your shoulders. Simply put, you must compute the volume of the intersection of anywhere from 2 to 1000 cubes.

Input
The input consists of several sets of cubes for which the volume of their intersections must be computed. The first line of each set contains a number (from 2 to 1000) which indicates the number of cubes which follow, one cube per line. Each line which describes a cube contains four integers. The first three integers are the x, y, and z coordinates of the corner of a cube, and the fourth integer is the positive distance which the cube extends in each of the three directions (parallel to the x, y, and z axes) from that corner.

Following the data for the first set of cubes will be a number which indicates how many cubes are in a second set, followed by the cube descriptions for the second set, again one per line. Following this will be a third set, and so on.A number =0 indicate the end of input.

Output
Your program should process sets all of cubes, outputting the volume of their intersections to the output file, one set per line, until a zero is read for the number of cubes.

Sample Input

 

Sample Output

 

Source
Mid-Central USA 1993

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Source Code

#include <iostream>
using namespace std;
struct Pot {
	int x,y,z;
}L,R,C;
int main() {
	int N,x,y,z,len;
	while(cin >> N,N!=0) {
		L.x = L.y = L.z = len = 0;
		cin >> L.x >> L.y >> L.z >> len;
		R.x = L.x + len;
		R.y = L.y + len;
		R.z = L.z + len;
		N--;
		bool flag = 1;
		while(N--) {
			C.x = C.y = C.z = len = 0;
			cin >> C.x >> C.y >> C.z >> len;
			if(C.x>=R.x||C.y>=R.y||C.z>=R.z)	{	flag = 0;break;}
			if(C.x+len<=L.x||C.y+len<=L.y||C.z+len<=L.z)	{flag =  0;break;}
			if(C.x>L.x)	L.x = C.x;
			if(C.y>L.y)	L.y = C.y;
			if(C.z>L.z)	L.z = C.z;
			if(C.x+len<R.x)	R.x = C.x+len;	
			if(C.y+len<R.y)	R.y = C.y+len;
			if(C.z+len<R.z)	R.z = C.z+len;
		}
		while(N>0)	{N--;cin >> C.x >> C.y >> C.z >> len;}
		if(flag == 0)	{cout << 0 << endl; continue;}
		cout << (R.x-L.x)*(R.y-L.y)*(R.z-L.z) << endl;
	}
	return 0;
}

25
9

2
0 0 0 10
9 1 1 5
3
0 0 0 10
9 1 1 5
8 2 2 3
0