最近考完试了。闲下来就正好刷刷以前没刷完的搜索专题。
简单搜索就没啥好讲的啦。就是暴力bfs和dfs。
这篇博客是kuangbin搜索进阶的专题的总结
八数码问题太经典啦。通过它来学习搜索的进阶技巧就很舒服。
首先是最简单的康拓优化。
康托展开（Hash序列权值,可以用于序列打表：
int jc[10] = {1,1,2,6,24,120,720,5040,40320,362880}; int cantor(int a[])//康托展开 { int ans = 0, k; for (int i = 0; i < 9; i++) { k=0; for (int j = i+1; j < 9; j++) if (a[i] > a[j]) k++; ans += k * jc[8-i]; } return ans; }
关于八数码的解法：
1.康托展开+逆向BFS预处理
#includeusing namespace std; const int N = 400000; struct path { char oper; int fa; }p[N]; int jc[10] = {1,1,2,6,24,120,720,5040,40320,362880}; int in[9]; int goal[]={1,2,3,4,5,6,7,8,9}; int vis[N]; int dic[4][2] = {1,0, -1,0, 0,1, 0,-1}; struct node { int maze[9]; int num; }; int cantor(int a[])//康托展开 { int ans = 0, k; for (int i = 0; i < 9; i++) { k = 0; for (int j = i+1; j < 9; j++) if (a[i] > a[j]) k++; ans += k * jc[8-i]; } return ans; } void bfs(int qin[]) { queue q; vis[cantor(qin)] = 1; node st; for (int i = 0; i < 9; i++) st.maze[i] = qin[i]; p[0].fa = -1; st.num = 8; q.push(st); while (!q.empty()) { node top = q.front(); q.pop(); for (int i = 0; i < 4; i++) { int tx = top.num % 3 + dic[i][0]; int ty = top.num / 3 + dic[i][1]; if (tx < 0) continue; if (tx >= 3) continue; if (ty < 0) continue; if (ty >= 3) continue; node temp; memcpy(temp.maze, top.maze, sizeof(top.maze)); temp.num = ty*3+tx; swap(temp.maze[top.num], temp.maze[temp.num]); int ct2 = cantor(temp.maze); int ct1 = cantor(top.maze); if (!vis[ct2]) { vis[ct2] = 1; q.push(temp); if (i == 0) p[ct2].fa = ct1, p[ct2].oper = 'l'; else if (i == 1) p[ct2].fa = ct1, p[ct2].oper = 'r'; else if (i == 2) p[ct2].fa = ct1, p[ct2].oper = 'u'; else if (i == 3) p[ct2].fa = ct1, p[ct2].oper = 'd'; //cout << p[ct2].oper << endl; } } } } int main() { cin.tie(0); ios::sync_with_stdio(0); bfs(goal); char temp[9]; while (cin >> temp[0] >> temp[1] >> temp[2] >> temp[3] >> temp[4] >> temp[5] >> temp[6] >> temp[7] >> temp[8]) { for (int i = 0; i < 9; i++) if (temp[i] == 'x') in[i] = 9; else in[i] = temp[i]-'0'; int cont = cantor(in); if (!vis[cont]) cout << "unsolvable" << endl; else { while (p[cont].fa != -1) { cout << p[cont].oper; cont = p[cont].fa; } cout << endl; } } return 0; }
2.双向BFS+康托展开+逆序对判断（我之前是拓展再判断结果wa了好久，改成先判断再拓展就一发过了真的服了
#include#include #include #include <string> #include #include #include using namespace std; const int N = 400000; struct path { char oper; int fa; }p_st[N], p_ed[N]; struct node { int maze[9]; int num; }st, ed; int jc[10] = {1,1,2,6,24,120,720,5040,40320,362880}; int in[9], goal[9] = {1,2,3,4,5,6,7,8,9}; int st_vis[N], ed_vis[N]; int st_mp[N], ed_mp[N]; int st_out, ed_out; int dic[4][2] = {1,0, -1,0, 0,1, 0,-1}; void output1(int k) { if(p_st[k].fa==-1) return ; else { output1(p_st[k].fa); cout << p_st[k].oper; } } void output2(int k) { if(p_ed[k].fa==-1) return ; else { //cout << k << endl; cout << p_ed[k].oper; output2(p_ed[k].fa); } } int cantor(int a[]) { int ans = 0, k; for (int i = 0; i < 9; i++) { k = 0; for (int j = i+1; j < 9; j++) if (a[i] > a[j]) k++; ans += k * jc[8-i]; } return ans; } void dbfs() { memset(st_vis, 0, sizeof(st_vis)); memset(ed_vis, 0, sizeof(ed_vis)); queue q1, q2; for (int i = 0; i < 9; i++) st.maze[i] = in[i], ed.maze[i] = goal[i]; p_ed[0].fa = -1; ed_vis[0] = 1; ed.num = 8; p_st[cantor(st.maze)].fa = -1; st_vis[cantor(st.maze)] = 1; q1.push(st); q2.push(ed); while (!q1.empty() && !q2.empty()) { node st_top = q1.front(); q1.pop(); int st_cur = cantor(st_top.maze); if (ed_vis[st_cur]) { st_out = st_cur; ed_out = st_cur; return ; } for (int i = 0; i < 4; i++) { int tx = st_top.num % 3 + dic[i][0]; int ty = st_top.num / 3 + dic[i][1]; if (tx < 0) continue; if (tx >= 3) continue; if (ty < 0) continue; if (ty >= 3) continue; node st_temp; memcpy(st_temp.maze, st_top.maze, sizeof(st_top.maze)); st_temp.num = ty*3+tx; swap(st_temp.maze[st_top.num], st_temp.maze[st_temp.num]); int aft = cantor(st_temp.maze); if (!st_vis[aft]) { st_vis[aft] = 1; q1.push(st_temp); if (i == 0) p_st[aft].fa = st_cur, p_st[aft].oper = 'r'; else if (i == 1) p_st[aft].fa = st_cur, p_st[aft].oper = 'l'; else if (i == 2) p_st[aft].fa = st_cur, p_st[aft].oper = 'd'; else if (i == 3) p_st[aft].fa = st_cur, p_st[aft].oper = 'u'; } } node ed_top = q2.front(); q2.pop(); int ed_cur = cantor(ed_top.maze); if (st_vis[ed_cur]) { st_out = ed_cur; ed_out = ed_cur; return ; } for (int i = 0; i < 4; i++) { int tx = ed_top.num % 3 + dic[i][0]; int ty = ed_top.num / 3 + dic[i][1]; if (tx < 0) continue; if (tx >= 3) continue; if (ty < 0) continue; if (ty >= 3) continue; node ed_temp; memcpy(ed_temp.maze, ed_top.maze, sizeof(ed_top.maze)); ed_temp.num = ty*3+tx; swap(ed_temp.maze[ed_top.num], ed_temp.maze[ed_temp.num]); int aft = cantor(ed_temp.maze); if (!ed_vis[aft]) { ed_vis[aft] = 1; q2.push(ed_temp); if (i == 0) p_ed[aft].fa = ed_cur, p_ed[aft].oper = 'l'; else if (i == 1) p_ed[aft].fa = ed_cur, p_ed[aft].oper = 'r'; else if (i == 2) p_ed[aft].fa = ed_cur, p_ed[aft].oper = 'u'; } } } } int main() { char temp[9]; while (cin >> temp[0] >> temp[1] >> temp[2] >> temp[3] >> temp[4] >> temp[5] >> temp[6] >> temp[7] >> temp[8]) { for (int i = 0; i < 9; i++) if (temp[i] == 'x') in[i] = 9, st.num = i; else in[i] = temp[i]-'0'; int rev = 0; for (int i = 0; i < 9; i++) for (int j = i+1; j < 9; j++) if (in[i] == 9) continue; else if (in[i] > in[j]) rev++; if (rev & 1) cout << "unsolvable" << endl; else { dbfs(); output1(st_out); output2(ed_out); cout << endl; } } return 0; }
这里顺便在总结一下DFBS(双向bfs)的要点：
1.两个queue, 两个path(记录父路径), 两个vis（用于检测碰撞与判重），两个输出的函数 PS: 别忘了初始化
2.一定是先判断之后再拓展路径，而不是反过来