Area of Polycubes(简单模拟)

来源：互联网发布：thinkpad x系列知乎编辑：程序博客网时间：2024/05/29 21:28

http://poj.org/problem?id=3792

题意：给出n个立方体的三维坐标，在每个立方体与之前的立方体有公共边的前提下输出他们的表面积，否则输出NO，并输出不合法的立方体编号。

注意有重的立方体，也输出NO。

#include <stdio.h>#include <string.h>#include <algorithm>#include <cmath>using namespace std;struct node{int x,y,z;}p[110];int main(){int test;int n;scanf("%d",&test);for(int item = 1; item <= test; item++){scanf("%d",&n);for(int i = 1; i <= n; i++){scanf("%d,%d,%d",&p[i].x,&p[i].y,&p[i].z);}int ok = 1;int ans = 6*n;printf("%d ",item);for(int i = 2; i <= n; i++){int cnt = 0;for(int j = 1; j < i; j++){if(p[i].x == p[j].x && p[i].y == p[j].y && p[i].z == p[j].z){ok = 0;printf("NO %d\n",i);break;}if( (p[i].x == p[j].x && p[i].y == p[j].y && abs(p[i].z-p[j].z) == 1) ||(p[i].x == p[j].x && p[i].z == p[j].z && abs(p[i].y-p[j].y) == 1) ||(p[i].y == p[j].y && p[i].z == p[j].z && abs(p[i].x-p[j].x) == 1) ){ans -=2;cnt++;}}if(ok == 0) break;if(cnt == 0){printf("NO %d\n",i);ok = 0;break;}}if(ok)printf("%d\n",ans);}return 0;}

0 0