The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
InputYou will receive, several descriptions of configuration of the 8 puzzle. One description is just a list of the tiles in their initial positions, with the rows listed from top to bottom, and the tiles listed from left to right within a row, where the tiles are represented by numbers 1 to 8, plus 'x'. For example, this puzzle
1 2 3
x 4 6
7 5 8
is described by this list:
1 2 3 x 4 6 7 5 8
OutputYou will print to standard output either the word ``unsolvable'', if the puzzle has no solution, or a string consisting entirely of the letters 'r', 'l', 'u' and 'd' that describes a series of moves that produce a solution. The string should include no spaces and start at the beginning of the line. Do not print a blank line between cases.
Sample Input
2 3 4 1 5 x 7 6 8
Sample Output
ullddrurdllurdruldr
题意：给出8个数和一个空位（x表示），空位可以向四个方向和数交换，求最小的操作方式试其变成 （1 2 3 4 5 6 7 8 x）
思路：奇数码问题，先判断是否有解，因为是奇数码，所以有解的情况是两个情况的逆序对奇偶相同。
另外，可以把x置换成9，然后对这个9位数进行标记，可以用map<int,int>,也可以用康托展开将其和自然数一一对应。
A*的评估函数是每个位置到其应当位置的曼哈顿距离
A* + map + string路径记录：
#include<iostream> #include<cstdio> #include<queue> #include<algorithm> #include<string.h> #include<map> using namespace std; struct Node { int cost,status; int s[10]; int pos; int f; string ans; Node(int co=0,int st=0,int pos=0,int f=0,string ans=""):cost(co),status(st),pos(pos),f(f),ans(ans) {} bool operator <(const Node x)const { return cost > x.cost; } }; map<int,int>mp; int ways[4][2] = {1,0,-1,0,0,1,0,-1}; char WAYS[4] = {'d','u','r','l'}; int ending; Node start; string ans; int Merge(int *a,int l,int r) { int mid = (r+l)>>1; int i=l,j=mid+1; int b[r-l+5]; int cnt = 0; int k = 0; while(i <= mid && j <= r) { if(a[i] <= a[j]) { b[cnt++] = a[i++]; } else { b[cnt++] = a[j++]; k += mid - i + 1; } } while(i <= mid) b[cnt++] = a[i++]; while(j <= r) b[cnt++] = a[j++]; for(int i=0; i<cnt; i++) { a[l++] = b[i]; } return k; } void MER(int *a,int l,int r,int &cnt) { if(l >= r) return; int mid = (l+r)>>1; MER(a,l,mid,cnt); MER(a,mid+1,r,cnt); cnt += Merge(a,l,r); } int turn(int s[]) { int cnt = 0; for(int i=1; i<=9; i++) { cnt *= 10; cnt += s[i]; } return cnt; } int init(char word[]) { int a[10]; for(int i=2; i<=9; i++) scanf(" %c",&word[i]); int cnt = 0; for(int i=1; i<=9; i++) if((word[i]) != 'x') a[++cnt] = word[i]-'0'; int num = 0; MER(a,1,cnt,num); ending = 0; for(int i=1; i<=9; i++) { ending *= 10; ending += i; if(word[i] == 'x') start.s[i] = 9,start.pos = i; else start.s[i] = word[i] - '0'; } return (num & 1) == 1; } int cal(int s[]) { int cnt = 0; for(int i=1; i<=9; i++) { if(s[i] == 9) continue; int x=(i-1)/3+1,y=(i-1)%3+1; int xx=(s[i]-1)/3+1,yy=(s[i]-1)%3+1; cnt += abs(x-xx)+abs(y-yy); } return cnt; } bool A_bfs(Node start) { priority_queue<Node>que; while(!que.empty()) que.pop(); start.status = turn(start.s); start.f = cal(start.s); start.cost = start.f; que.push(start); while(!que.empty()) { Node tmp = que.top(); que.pop(); int status = tmp.status; int cost = tmp.cost - tmp.f; if(mp[status]) continue; mp[status] = 1; if(status == ending) { ans = tmp.ans; return 1; }; int x = (tmp.pos-1)/3+1; int y = (tmp.pos-1)%3+1; for(int i=0; i<4; i++) { int xx = x+ways[i][0]; int yy = y+ways[i][1]; if(xx < 1 || xx > 3 || yy < 1 || yy > 3) continue; Node t = tmp; t.s[(x-1)*3+y] = tmp.s[(xx-1)*3+yy]; t.s[(xx-1)*3+yy] = 9; t.status = turn(t.s); t.f = cal(t.s); t.cost = cost+1+t.f; t.pos = (xx-1)*3+yy; t.ans += WAYS[i]; que.push(t); } } return 0; } int main() { char word[10]; while(~scanf(" %c",&word[1])) { mp.clear(); if(init(word)) printf("unsolvable\n"); else { A_bfs(start); cout << ans << endl; } } }
双向bfs+康拓展开+递归路径记录（注：路径记录不能都用递归，会ME，也许姿势不对，对于从终态搜索的，因为其本身就是倒序，可以采用递推，从当前状态推过去）
#include<bits/stdc++.h> using namespace std; struct Node { int cost,status; int s[10]; int pos; Node(int co=0,int st=0,int pos=0):cost(co),status(st),pos(pos){} }; const int maxn = 4e5; int fac[9] = {1,1,2,6,24,120,720,5040,40320}; int ways[4][2] = {1,0,-1,0,0,1,0,-1}; char WAYS[4] = {'d','u','r','l'}; char FWAYS[4]= {'u','d','l','r'}; char path[maxn],path2[maxn]; int pre[maxn],pre2[maxn]; bool vis[2][maxn]; int ans; queue<Node>que[2]; Node start,ending; char word[10]; int a[10],b[10]; int cantor(int s[]) { int sum = 0; for(int i=1; i<=9; i++) { int num = 0; for(int j=i+1; j<=9; j++) { if(s[j] < s[i]) num++; } sum += num*fac[9-i]; } return sum+1; } int Merge(int *a,int l,int r) { int mid = (r+l)>>1; int i=l,j=mid+1; int cnt = 0; int k = 0; while(i <= mid && j <= r) { if(a[i] <= a[j]) { b[cnt++] = a[i++]; } else { b[cnt++] = a[j++]; k += mid - i + 1; } } while(i <= mid) b[cnt++] = a[i++]; while(j <= r) b[cnt++] = a[j++]; for(int i=0; i<cnt; i++) { a[l++] = b[i]; } return k; } void MER(int *a,int l,int r,int &cnt) { if(l >= r) return; int mid = (l+r)>>1; MER(a,l,mid,cnt); MER(a,mid+1,r,cnt); cnt += Merge(a,l,r); } int init(char word[]) { memset(vis,0,sizeof(vis)); for(int i=2; i<=9; i++) scanf(" %c",&word[i]); int cnt = 0; for(int i=1; i<=9; i++) if((word[i]) != 'x') a[++cnt] = word[i]-'0'; int num = 0; MER(a,1,cnt,num); for(int i=1; i<=9; i++) { ending.s[i] = i; if(word[i] == 'x') start.s[i] = 9,start.pos = i; else start.s[i] = word[i] - '0'; } start.status = cantor(start.s); ending.status = cantor(ending.s); ending.pos = 9; return (num & 1) == 1; } bool bfs(int w) { int num = que[w].size(); while(num--) { Node tmp = que[w].front(); que[w].pop(); int status = tmp.status; int cost = tmp.cost; if(vis[w^1][status]) { ans = status; return 1; } int x = (tmp.pos-1)/3+1; int y = (tmp.pos-1)%3+1; for(int i=0; i<4; i++) { int xx = x+ways[i][0]; int yy = y+ways[i][1]; if(xx < 1 || xx > 3 || yy < 1 || yy > 3 ) continue; Node t = tmp; t.s[(x-1)*3+y] = tmp.s[(xx-1)*3+yy]; t.s[(xx-1)*3+yy] = 9; t.status = cantor(t.s); if(vis[w][t.status])continue; vis[w][t.status] = 1; t.s[(x-1)*3+y] = tmp.s[(xx-1)*3+yy]; t.s[(xx-1)*3+yy] = 9; t.cost = cost+1; t.pos = (xx-1)*3+yy; if(w == 0)path[t.status] = WAYS[i],pre[t.status] = status; else path2[t.status] = FWAYS[i],pre2[t.status] = status; que[w].push(t); } } return 0; } void out(int w) { if(pre[w] == -1) return; out(pre[w]); printf("%c",path[w]); } void out2(int w) { while(pre2[w] != -1) { printf("%c",path2[w]); w = pre2[w]; } } void solve() { while(!que[0].empty())que[0].pop(); while(!que[1].empty())que[1].pop(); que[0].push(start); que[1].push(ending); pre[start.status] = pre2[ending.status] = -1; vis[0][start.status] = vis[1][ending.status] = 1; while(!que[0].empty() || !que[1].empty()) { if(bfs(0)) { out(ans); out2(ans); puts(""); return; } if(bfs(1)) { out(ans); out2(ans); puts(""); return; } } } int main() { while(~scanf(" %c",&word[1])) { if(init(word)) printf("unsolvable\n"); else { solve(); } } }