Ultraviolet A 810-a dicey problem (BFS)

Ultraviolet A 810-a dicey Problem

Question Link

Question: a dice, giving you the top and the front, at a starting point, each time you can move to the four cells around, for-1, or the top is the same as the number at this position, then the problem arises, whether the dice can go back to the original place in one lap. The output path must be the shortest. If there are multiple shortest paths, output the results based on the upper, lower, and lower paths.

Train of Thought: to understand the question is a water question, you can simply perform a memory-based search, record the status as position and dice top, Front (because there are two sides to determine the dice)

Code:

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int D[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};const int N = 15;char str[25];int n, m, sx, sy, u, f, to[7][7], g[N][N];int vis[N][N][7][7];struct State {int x, y, u, f;int pre;State() {}State(int x, int y, int u, int f, int pre) {this->x = x;this->y = y;this->u = u;this->f = f;this->pre = pre;}} Q[10005];void tra(int &vu, int &vf, int d) {if (d == 0) {int tmp = vf; vf = 7 - vu; vu = tmp;}if (d == 1) {int tmp = vu; vu = 7 - vf; vf = tmp;}if (d == 2) vu = 7 - to[vu][vf];if (d == 3) vu = to[vu][vf];}#define MP(a,b) make_pair(a,b)typedef pair<int, int> pii;vector<pii> ans;void print(int u) {if (u == -1) return;print(Q[u].pre);ans.push_back(MP(Q[u].x, Q[u].y));}void bfs() {ans.clear();int head = 0, rear = 0;Q[rear++] = State(sx, sy, u, f, -1);memset(vis, 0, sizeof(vis));vis[sx][sy][u][f] = 1;while (head < rear) {State u = Q[head++];for (int i = 0; i < 4; i++) {State v = u;v.x += D[i][0];v.y += D[i][1];if (v.x <= 0 || v.x > n || v.y <= 0 || v.y > m) continue;if (g[v.x][v.y] != -1 && u.u != g[v.x][v.y]) continue;if (v.x == sx && v.y == sy) {print(head - 1);ans.push_back(MP(sx, sy));int tot = ans.size();for (int i = 0; i < tot; i++) {if (i % 9 == 0) printf("\n  ");printf("(%d,%d)%c", ans[i].first, ans[i].second, i == tot - 1 ? '\n' : ',');}return;}tra(v.u, v.f, i);if (vis[v.x][v.y][v.u][v.f]) continue;vis[v.x][v.y][v.u][v.f] = 1;v.pre = head - 1;Q[rear++] = v;}}printf("\n  No Solution Possible\n");}int main() {to[1][2] = 4; to[1][3] = 2; to[1][4] = 5; to[1][5] = 3;to[2][1] = 3; to[2][3] = 6; to[2][4] = 1; to[2][6] = 4;to[3][1] = 5; to[3][2] = 1; to[3][5] = 6; to[3][6] = 2;to[4][1] = 2; to[4][2] = 6; to[4][5] = 1; to[4][6] = 5;to[5][1] = 4; to[5][3] = 1; to[5][4] = 6; to[5][6] = 3;to[6][2] = 3; to[6][3] = 5; to[6][4] = 2; to[6][5] = 4;while (~scanf("%s", str) && strcmp(str, "END")) {printf("%s", str);scanf("%d%d%d%d%d%d", &n, &m, &sx, &sy, &u, &f);for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)scanf("%d", &g[i][j]);bfs();}return 0;}

Ultraviolet A 810-a dicey problem (BFS)