UVA 810 - A Dicey Problem(BFS)
UVA 810 - A Dicey Problem
题意:一个骰子,给你顶面和前面。在一个起点,每次能移动到周围4格,为-1,或顶面和该位置数字一样,那么问题来了,骰子能不能走一圈回到原地,输出路径,要求最短,假设有多个最短,依照上下左右输出
思路:读懂题就是水题,就记忆化搜一下就可以,记录状态为位置和骰子顶面。正面(由于有两面就能确定骰子了)
代码:
#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; }
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