HDOJ 1043 康拓展开

Eight

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26403    Accepted Submission(s): 7018
Special Judge

Problem 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 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.

 

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
  
  

 

Source

South Central USA 1998 (Sepcial Judge Module By JGShining)

 

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代码如下：

#include <cstdio>
#include <queue>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1e6+100;
int fac[15];
struct node{
    int x,y;
    int cnt;
    char num[9];
};
bool vis[maxn];
int dx[]={0,0,-1,1};
int dy[]={1,-1,0,0};
char path[maxn];
int pre[maxn];
queue<struct node> que;

void init(){
   fac[0] = fac[1] = 1;
   for (int i=2; i<=10; i++) fac[i] = fac[i-1]*i;
}

int contour(char *s){
     int sum = 0,cnt;
     for (int i=0; i<8; i++) {
         cnt = 0;
         for (int j=i+1; j<9; j++) if (s[i]>s[j]) cnt++;
         sum += cnt*fac[9-i-1];
     }
     return sum+1;
}

void BFS(){
    while (!que.empty()) que.pop();
    memset(vis,0,sizeof(vis));

    node p,q;
    for (int i=0; i<8; i++) p.num[i] = i+1+'0';
    p.num[8] = '0';
    p.cnt = contour(p.num);
    p.x = p.y = 2;
    vis[p.cnt] = 1;
    path[p.cnt] = -1;

    que.push(p);
    while (!que.empty()){
         p = que.front();
         que.pop();
         for (int i=0; i<4; i++) {
            q.x = p.x+dx[i];
            q.y = p.y+dy[i];
            if (q.x<0 || q.y<0 || q.x>2 || q.y>2) continue;

            int pos = q.x*3+q.y;
            int p_pos = p.x*3+p.y;

            strcpy(q.num,p.num);
            swap(q.num[p_pos],q.num[pos]);
            q.cnt = contour(q.num);

            if (!vis[q.cnt]){
                vis[q.cnt] = 1;
                pre[q.cnt] = p.cnt;

                switch (i){
                    case 0: path[q.cnt] = 'l'; break;
                    case 1: path[q.cnt] = 'r'; break;
                    case 2: path[q.cnt] = 'd'; break;
                    default: path[q.cnt] = 'u';
                }

                que.push(q);
            }
         }
    }

}

char str[100];
char cc[10];
int tot;

void Print(int f){
    int i=f;
    while (path[i]!=-1) {
        printf("%c",path[i]);
        i = pre[i];
    }
    printf("\n");
}

int main(){
    init();
    BFS();
    while (gets(str)!=NULL){
        int len = strlen(str);
        tot = 0;
        for (int i=0; i<len; i++) {
             if (str[i]=='x') cc[tot++] = '0';
             else if (str[i]>='1' && str[i]<='8') cc[tot++] = str[i];
        }
        cc[tot] = '\0';
        int f = contour(cc);
        if (!vis[f]) printf("unsolvable\n");
        else Print(f);
    }
    return 0;
}