ZOJ - 2477 dfs [kuangbin带你飞]专题二

来源：互联网发布：dnf 登陆器源码编辑：程序博客网时间：2024/05/29 18:24

注意输入的处理，旋转操作打表。递增枚举可能步数，作为限制方便找到最短路。

AC代码：90ms

#include<cstdio>#include<cstring>char magic[60];int cube[6][9] = {{4,0,1,2,3,5,6,7,8},{22,9,10,11,21,23,33,34,35},{25,12,13,14,24,26,36,37,38},    {28,15,16,17,27,29,39,40,41},{31,18,19,20,30,32,42,43,44},{49,45,46,47,48,50,51,52,53}}; const int rot[12][20]={         {11,23,35,34,33,21, 9,10,51,48,45,36,24,12, 6, 3, 0,20,32,44}, //只有20个需要改变，每个中心无需改变          { 9,10,11,23,35,34,33,21,36,24,12, 6, 3, 0,20,32,44,51,48,45},         {14,13,26,38,37,36,24,12,45,46,47,39,27,15, 8, 7, 6,11,23,35},         {12,24,13,14,26,38,37,36,39,27,15, 8, 7, 6,11,23,35,45,46,47},         {17,29,41,40,39,27,15,16,47,50,53,42,30,18, 2, 5, 8,14,26,38},         {15,16,17,29,41,40,39,27,42,30,18, 2, 5, 8,14,26,38,47,50,53},         {18,19,20,32,44,43,42,30,53,52,51,33,21, 9, 0, 1, 2,17,29,41},         {42,30,18,19,20,32,44,43,33,21, 9, 0, 1, 2,17,29,41,53,52,51},         { 0, 1, 2, 5, 8, 7, 6, 3,12,13,14,15,16,17,18,19,20, 9,10,11},         { 6, 3, 0, 1, 2, 5, 8, 7,15,16,17,18,19,20, 9,10,11,12,13,14},         {45,46,47,50,53,52,51,48,44,43,42,41,40,39,38,37,36,35,34,33},         {51,48,45,46,47,50,53,52,41,40,39,38,37,36,35,34,33,44,43,42}};bool is_ok() {for(int i = 0; i < 6; ++i) {for(int j = 0; j < 8; ++j) {if(magic[cube[i][j]] != magic[cube[i][j+1]]) return false;}}return true;}void rotate(int k) { //旋转第k面 int h = k ^ 1; //对应转动方式,顺时针or逆时针 char tmp[60];memcpy(tmp, magic, sizeof(magic));for(int i = 0; i < 20; ++i) {magic[rot[k][i]] = tmp[rot[h][i]];}}int ans[2][10];int limit;int dfs(int cnt) {if(cnt >= limit) return is_ok();char old[60];memcpy(old, magic, sizeof(magic));for(int i = 0; i < 12; ++i) {rotate(i);ans[0][cnt] = i / 2;ans[1][cnt] = (i & 1) ? -1 : 1;if(dfs(cnt + 1)) return 1;memcpy(magic, old, sizeof(old));}} int main() {int T;scanf("%d", &T);char str[60];int kase = 0;while(T-- ) {if(kase++) getchar();for(int i = 0; i < 54; ++i) {while(1) {char ch = getchar();if(ch >= 'a' && ch <= 'z') {magic[i] = ch;break;}}}for(limit = 0; limit <= 6; ++limit) {if(limit == 6) printf("-1\n");else if(dfs(0)) {printf("%d\n", limit);for(int i = 0; i < limit; ++i) {printf("%d %d\n", ans[0][i], ans[1][i]);}break;}}}return 0;}

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