題目鏈接:
Image Is Everything
解析:
首先假設該正方體是滿的,之後基於以下規則迭代:
- 如果視圖爲’.’,那麼該面下面的所有立方體都要刪除。
- 在遍歷六視圖進行判斷的時候,如果該面沒有塗上顏色, 那麼我們就假設這個面是表面,把他塗上顏色即可。
- 如果該面已塗的顏色和當前六視圖對應面的顏色相同,即不存在矛盾,那麼繼續判斷六視圖下一個面。否則,存在矛盾,該立方體刪除。
AC代碼:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define RED(i,n) for(int i = 0; i < (n); i++)
#define MAXN (10+5)
int n;
char view[6][MAXN][MAXN], pos[MAXN][MAXN][MAXN];
char read_char(){
char c;
int t = 0;
while(true){
scanf("%c", &c);
if(c == '.' || (c >= 'A' && c <= 'Z')) return c;
}
}
void get(int t, int i, int j, int len, int &x, int &y, int &z){
if(!t){x = len; y = j; z = i; return;}
if(t == 1){x = n-j-1; y = len; z = i; return;}
if(t == 2){x = n-len-1; y = n-j-1; z = i; return;}
if(t == 3){x = j; y = n-len-1; z = i; return;}
if(t == 4){x = n-i-1; y = j; z = len; return;}
if(t == 5){x = i; y = j; z = n-len-1; return;}
}
int main(){
while(scanf("%d", &n) != EOF && n){
RED(i,n) RED(j,6) RED(k,n) view[j][i][k] = read_char();
RED(i,n) RED(j,n) RED(k,n) pos[i][j][k] = '#';
RED(k,6) RED(i,n) RED(j,n)
if(view[k][i][j] == '.'){
RED(p, n){
int x, y, z;
get(k, i, j, p, x, y, z);
pos[x][y][z] = '.';
}
}
while(true){
bool ok = true;
RED(k,6) RED(i,n) RED(j,n){
if(view[k][i][j] != '.'){
RED(p,n){
int x, y, z;
get(k, i, j, p, x, y, z);
if(pos[x][y][z] == '.') continue;
if(pos[x][y][z] == '#'){
pos[x][y][z] = view[k][i][j];
break;
}
if(pos[x][y][z] == view[k][i][j]) break;
ok = false;
pos[x][y][z] = '.';
}
}
}
if(ok) break;
}
int ans = 0;
RED(i,n) RED(j,n) RED(k,n)
if(pos[i][j][k] != '.') ans++;
printf("Maximum weight: %d gram(s)\n", ans);
}
return 0;