八数码--HDU - 1043 Eight

Description

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: 

 1  2  3  4 5  6  7  8 9 10 11 1213 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 1213 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. 

 

Input

You 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 

 

Output

You 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 

分析：典型的八数码问题，整个搜索过程看成图的遍历，每个数码状态即图的结点，采用bfs遍历求最短路；要注意数码状态的保存可以用hash，标准库会超时；另一种方法是逆向bfs计算所有可能的状态，然后用输入判断遍历结果：

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;const int maxn = 500000;const int dr[] = { -1, 0, 1, 0 };const int dc[] = { 0, 1, 0, -1 };const char chd[] = {'u', 'r', 'd', 'l'};const int goal[] = { 1, 2, 3, 4, 5, 6, 7, 8, 0 };typedef int State[9];State st[maxn], src;int parent[maxn], id[maxn], dis[maxn], vis[362880], fact[9];char dir[maxn], ans[maxn];void print_path(int u){vector<char> ans;while(u != 1) {ans.push_back(dir[u]);u = parent[u];}for(int i=0; i<ans.size(); i++) putchar(ans[i]);putchar('\n');}bool unvisit(int s){int sum = 0;for(int i=0; i<9; i++) {int cnt = 0;for(int j=i+1; j<9; j++) if(st[s][j]<st[s][i]) cnt++;sum += fact[8-i]*cnt;}if(vis[sum]) return false;id[sum] = s;return vis[sum]=1;}int bfs(){memset(vis, 0, sizeof(vis));int front = 1, rear = 2;unvisit(1); dis[front] = 0;while(front < rear) {State& s = st[front];//if(memcmp(s, goal, sizeof(goal)) == 0) return front;int z;        for(z=0; z<9; z++) if(!s[z]) break;        int x = z/3, y = z%3;        for(int d=0; d<4; d++) {int newx = x + dr[d];int newy = y + dc[d];int newz = newx*3 + newy;if(newx>=0&&newx<3 && newy>=0&&newy<3) {State& t = st[rear];                memcpy(&t, &s, sizeof(s));                t[newz] = s[z];                t[z] = s[newz];dir[rear] = chd[(d+2)%4];                if(unvisit(rear)) {dis[rear] = dis[front] + 1;parent[rear] = front;rear++;                }}        }        front++;}return 0;}int main(){char tmp;fact[0] = 1;for(int i=1; i<9; i++) fact[i] = fact[i-1]*i;memcpy(st[1], goal, sizeof(goal));bfs();while(cin >> tmp) {if(tmp == 'x') tmp='0'; src[0] = tmp-'0';for(int i=1; i<9; i++) {cin >> tmp;if(tmp == 'x') src[i] = 0;else src[i] = tmp-'0';}        int sum = 0;        for(int i=0; i<9; i++) {int cnt = 0;for(int j=i+1; j<9; j++) if(src[j]<src[i]) cnt++;sum += fact[8-i]*cnt;        }        if(!vis[sum]) printf("unsolvable\n");        else print_path(id[sum]);}    return 0;}