八连块问题 紫书上的简单搜索 渣渣好久才弄懂
#include<cstdio> #include<cstring> using namespace std; const int M = 1000003; int x[4] = { -1, 1, 0, 0}, y[4] = {0, 0, -1, 1}; int dis[M], h[M], s[M][9], e[9]; int aton(int a[]) { int t = 0; for(int i = 0; i < 9; ++i) t = t * 10 + a[i]; return t; } int search_hash(int a[]) { int t = aton(a), p = t % M; while(h[p]) { if(h[p] == t) return p; ++p; if(p >= M) p = 0; } return p; } int bfs() { memset(h, 0, sizeof(h)); int fro = 1, rear = 2, r, c, k = 0, p; int t[9], tmp, cur, nr, nc, nk; while(fro < rear) { memcpy(t, s[fro], sizeof(t)); cur = search_hash(t); if(memcmp(t, e, sizeof(t)) == 0) return dis[cur]; for(k = 0; t[k];) ++k; r = k / 3, c = k % 3; for(int i = 0; i < 4; ++i) { memcpy(t, s[fro], sizeof(t)); nr = r + x[i], nc = c + y[i], nk = nr * 3 + nc; if(nr < 0 || nr > 2 || nc < 0 || nc > 2) continue; tmp = t[nk]; t[nk] = 0; t[k] = tmp; p = search_hash(t); if(h[p] == 0) { h[p] = aton(t); dis[p] = dis[cur] + 1; memcpy(s[rear], t, sizeof(t)); ++rear; } } ++fro; } return -1; } int main() { while(~scanf("%d", &s[1][0])) { memset(dis, 0, sizeof(dis)); for(int i = 1; i < 9; ++i) scanf("%d", &s[1][i]); for(int i = 0; i < 9; ++i) scanf("%d", &e[i]); h[aton(s[1]) % M] = aton(s[1]); int ans = bfs(); if(ans != -1) printf("%d\n", ans); else printf("No solution\n"); } return 0; } /* 2 6 4 1 3 7 0 5 8 8 1 5 7 3 6 4 0 2 2 3 4 1 5 0 7 6 8 1 2 3 4 5 6 7 8 0 */
时间: 2024-10-11 00:41:57