UVa 10024 - Curling up the cube

来源：互联网发布：mysql alias 编辑：程序博客网时间：2024/06/05 05:10

题目：给你一个6*6的平面，上面有一些方形（用1表示），问他们能不能拼成一个正方体。

分析：搜索、模拟。直接找到所有的可能的解，然后对平面进行旋转和翻转，匹配查找。

说明：uhunt和UVa最近都山不去啊。

所有的可能解如上图所示，直接打表匹配即可。

#include <cstdlib>#include <cstdio>int row[11] = {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2};int column[11] = {4, 4, 4 ,4 ,4 ,4 ,4 ,4 ,4 ,4 ,5};int block[11][3][5] = {1, 0, 0, 0, 0,  1, 1, 1, 1, 0,  1, 0, 0, 0, 0,1, 0, 0, 0, 0,  1, 1, 1, 1, 0,  0, 1, 0, 0, 0,1, 0, 0, 0, 0,  1, 1, 1, 1, 0,  0, 0, 0, 1, 0,1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0,0, 1, 0, 0, 0,  1, 1, 1, 1, 0,  0, 1, 0, 0, 0,0, 1, 0, 0, 0,  1, 1, 1, 1, 0,  0, 0, 1, 0, 0,1, 0, 0, 0, 0,  1, 1, 1, 0, 0,  0, 0, 1, 1, 0,0, 0, 1, 0, 0,  1, 1, 1, 0, 0,  0, 0, 1, 1, 0,0, 1, 0, 0, 0,  1, 1, 1, 0, 0,  0, 0, 1, 1, 0,1, 1, 0, 0, 0,  0, 1, 1, 0, 0,  0, 0, 1, 1, 0,1, 1, 1, 0, 0,  0, 0, 1, 1, 1,  0, 0, 0, 0, 0};int M[6][6], S[6][6];int main(){    int n;    while (~scanf("%d",&n))    while (n --) {for (int i = 0; i < 6; ++ i)for (int j = 0; j < 6; ++ j)    scanf("%d",&M[i][j]);int find = 0;for (int k = 0; k < 8; ++ k) {    for (int i = 0; i < 6; ++ i)    for (int j = 0; j < 6; ++ j)S[i][j] = M[i][j];    for (int t = 0; t < 11 && !find; ++ t)    for (int x = 0; x <= 6-row[t] && !find; ++ x)     for (int y = 0; y <= 6-column[t] && !find; ++ y) {int sum = 0;for (int p = 0; p < row[t]; ++ p)for (int q = 0; q < column[t]; ++ q)    if (S[x+p][y+q] && block[t][p][q])           sum ++;    if (sum == 6) {        find = 1;break;    }    }    if (find) break;    for (int i = 0; i < 6; ++ i)    for (int j = 0; j < 6; ++ j)M[i][j] = S[5-j][i];if (k == 3) {//翻转    for (int i = 0; i < 6; ++ i)    for (int j = 0; j < i; ++ j) {        M[i][j] ^= M[j][i];        M[j][i] ^= M[i][j];        M[i][j] ^= M[j][i];}}} if (find) puts("correct");else puts("incorrect");if (n) puts("");    }     return 0;}

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