题意:从无序到有序移动的方案,即最后成1 2 3 4 5 6 7 8 0
分析:八数码经典问题。POJ是一次,HDOJ是多次。因为康托展开还不会,也写不了什么,HDOJ需要从最后的状态逆向搜索,这样才不会超时。判重康托展开,哈希也可。
POJ
//#include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<string> #include<stack> #include<queue> #include <cstring> #include<map> #include<stdio.h> #include<stdlib.h> #include<ctype.h> #include<time.h> #include<math.h> using namespace std; const int N = 362880 + 5; const int MOD = 1e6 + 7; int dx[4] = {-1, 1, 0, 0}; int dy[4] = {0, 0, -1, 1}; char dir[4] = {‘u‘, ‘d‘, ‘l‘, ‘r‘}; struct Point { int s, d; string str; Point () {} Point (int s, int d, string str) : s (s), d (d), str (str) {} }; struct Hash_table { struct Edge { int v, nex; }edge[MOD]; int head[MOD], e; void init(void) { memset (head, -1, sizeof (head)); e = 0; } bool insert(int x) { int u = (x % MOD + MOD) % MOD; for (int i=head[u]; ~i; i=edge[i].nex) { if (edge[i].v == x) return false; } edge[e].v = x; edge[e].nex = head[u]; head[u] = e++; return true; } }ha; int vis[N], fact[9]; void decode(int x, int *b) { for (int i=8; i>=0; --i) { b[i] = x % 10; x /= 10; } } int encode(int *b) { int ret = 0; for (int i=0; i<9; ++i) { ret = ret * 10 + b[i]; } return ret; } int find_0(int *b) { for (int i=0; i<9; ++i) { if (b[i] == 0) return i; } return -1; } bool check(int x, int y) { if (x < 0 || x >= 3 || y < 0 || y >= 3) return false; else return true; } void print(int *b) { for (int i=0; i<9; ++i) { printf ("%d ", b[i]); if (i == 2 || i == 5 || i == 8) puts (""); } } void init(void) { fact[0] = 1; for (int i=1; i<9; ++i) fact[i] = fact[i-1] * i; memset (vis, false, sizeof (vis)); } bool can_insert(int *b) { int code = 0; for (int i=0; i<9; ++i) { int cnt = 0; for (int j=i+1; j<9; ++j) if (b[j] < b[i]) cnt++; code += fact[8-i] * cnt; } if (vis[code]) return false; else { vis[code] = true; return true; } } void BFS(int *a) { init (); int ans[9] = {1, 2, 3, 4, 5, 6, 7, 8, 0}; int s = encode (a); queue<Point> que; que.push (Point (s, 0, "")); while (!que.empty ()) { Point u = que.front (); que.pop (); int b[9]; decode (u.s, b); if (memcmp (ans, b, sizeof (b)) == 0) { int len = u.str.length (); for (int i=0; i<len; ++i) { printf ("%c", u.str[i]); } puts (""); return ; } int p = find_0 (b); int x = p / 3, y = p % 3; for (int i=0; i<4; ++i) { int tx = x + dx[i], ty = y + dy[i]; if (!check (tx, ty)) continue; int p2 = tx * 3 + ty; int t[9]; memcpy (t, b, sizeof (b)); t[p] = t[p2]; t[p2] = 0; int v = encode (t); //if (!ha.insert (v)) continue; if (!can_insert (t)) continue; que.push (Point (v, u.d + 1, u.str + dir[i])); } } puts ("unsolvable"); } int main(void) { char c[9]; int a[9]; while (scanf ("%c %c %c %c %c %c %c %c %c", &c[0], &c[1], &c[2], &c[3], &c[4], &c[5], &c[6], &c[7], &c[8]) == 9) { for (int i=0; i<9; ++i) { if (c[i] >= ‘1‘ && c[i] <= ‘9‘) { a[i] = c[i] - ‘0‘; } else a[i] = 0; } BFS (a); } return 0; }
HDOJ
#include <bits/stdc++.h> using namespace std; const int N = 362880 + 5; int dx[4] = {-1, 1, 0, 0}; int dy[4] = {0, 0, -1, 1}; char dir[4] = {‘d‘, ‘u‘, ‘r‘, ‘l‘}; struct Point { int s; int b[9]; }; struct Ans { char dir; int fa; }ans[N]; int fact[9]; int find_0(int *b) { for (int i=0; i<9; ++i) { if (b[i] == 9) return i; } return -1; } bool check(int x, int y) { if (x < 0 || x >= 3 || y < 0 || y >= 3) return false; else return true; } void init(void) { fact[0] = 1; for (int i=1; i<9; ++i) fact[i] = fact[i-1] * i; for (int i=0; i<N; ++i) ans[i].fa = -1; } int Cantor(int *b) { int code = 0; for (int i=0; i<9; ++i) { int cnt = 0; for (int j=i+1; j<9; ++j) if (b[j] < b[i]) cnt++; code += fact[8-i] * cnt; } return code; } void BFS() { init (); Point sta; for (int i=0; i<9; ++i) { sta.b[i] = i + 1; } sta.s = 0; ans[sta.s].fa = 0; queue<Point> que; que.push (sta); while (!que.empty ()) { Point u = que.front (); que.pop (); int p = find_0 (u.b); int x = p / 3, y = p % 3; for (int i=0; i<4; ++i) { Point v = u; int tx = x + dx[i], ty = y + dy[i]; if (!check (tx, ty)) continue; int p2 = tx * 3 + ty; swap (v.b[p], v.b[p2]); v.s = Cantor (v.b); if (ans[v.s].fa != -1) continue; ans[v.s].dir = dir[i]; ans[v.s].fa = u.s; que.push (v); } } } int main(void) { BFS (); char c[55]; int a[9]; while (gets (c)) { int j = 0; for (int i=0; c[i]; ++i) { if (c[i] >= ‘0‘ && c[i] <= ‘8‘) { a[j++] = c[i] - ‘0‘; } else if (c[i] == ‘x‘) a[j++] = 9; } int s = Cantor (a); if (ans[s].fa == -1) puts ("unsolvable"); else { while (s != 0) { printf ("%c", ans[s].dir); s = ans[s].fa; } puts (""); } } return 0; }
时间: 2024-10-12 14:23:03