UVA 10085 - The most distant state

题目链接：

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=110&page=show_problem&problem=1026

类型： 隐式图搜索

原题：

The 8-puzzle is a square tray in which eight square tiles are placed. The remaining ninth square is uncovered. Each tile has a number on it. A tile that is adjacent to the blank space can be slid into that space. A game consists of a starting state and a specified goal state. The starting state can be transformed into the goal state by sliding (moving) the tiles around. The 8-puzzle problem asks you to do the transformation in minimum number of moves.

 

2	8	3				1	2	3
1	6	4	=>			8		4
7		5				7	6	5
Start						Goal

 

However, our current problem is a bit different. In this problem, given an initial state of the puzzle your are asked to discover a goal state which is the most distant (in terms of number of moves) of all the states reachable from the given state.

Input

The first line of the input file contains an integer representing the number of test cases to follow. A blank line follows this line.

Each test case consists of 3 lines of 3 integers each representing the initial state of the puzzle. The blank space is represented by a 0 (zero). A blank line follows each test case.

 

Output

For each test case first output the puzzle number. The next 3 lines will contain 3 integers each representing one of the most distant states reachable from the given state. The next line will contain the shortest sequence of moves that will transform the given state to that state. The move is actually the movement of the blank space represented by four directions : U (Up), L (Left), D (Down) and R (Right). After each test case output an empty line.

 

Sample Input

1

2 6 4
1 3 7
0 5 8

Sample Output

Puzzle #1
8 1 5
7 3 6
4 0 2

UURDDRULLURRDLLDRRULULDDRUULDDR

题目大意：

题目都说了是八数码问题，不同的是，这个不是求一个八数码初始状态到达目标状态的最少步数， 而是要随便走，找出走的步数最多的那个状态，不能重复。

分析与总结：

解题过程与标准八数码问题基本相似， 不同的是，这个没有一个目标状态，所以要让它尽情地搜索，搜啊搜，一直到无法再搜下去了，那么最终状态就是步数最多的那个状态。在搜的过程记录下路径再输出即可。

// Time: 0.744s (UVA)
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#define MAXN 1000000
using namespace std;
char input[30];
int state[9], goal[9] = {1,2,3,8,0,4,7,6,5};
int dir[4][2] = {{-1,0},{1,0},{0,-1},{0,1}}; // 上，下，左， 右
char path_dir[5] = "UDLR";
int st[MAXN][9];
int father[MAXN], path[MAXN], ans; // 保存打印路径


const int MAXHASHSIZE = 1000003;
int head[MAXHASHSIZE], next[MAXN];

void init_lookup_table() { memset(head, 0, sizeof(head)); }

typedef int State[9];
int hash(State& s) {
  int v = 0;
  for(int i = 0; i < 9; i++) v = v * 10 + s[i];
  return v % MAXHASHSIZE;

}

int try_to_insert(int s) {
  int h = hash(st[s]);
  int u = head[h];
  while(u) {
    if(memcmp(st[u], st[s], sizeof(st[s])) == 0) return 0;
    u = next[u];
  }
  next[s] = head[h];
  head[h] = s;
  return 1;
}


void bfs(){
    
    init_lookup_table();
    father[0] = path[0] = -1;
    int front=0, rear=1;
    memcpy(st[0], state, sizeof(state));
    try_to_insert(0);

    while(front < rear){
        int *s = st[front];
       
        int j;
        for(j=0; j<9; ++j) if(!s[j])break; // 找出0的位置
        int x=j/3, y=j%3;     // 转换成行，列
        
        for(int i=0; i<4; ++i){

            int dx = x+dir[i][0]; // 新状态的行，列
            int dy = y+dir[i][1];
            int pos = dx*3+dy;    // 目标的位置

            if(dx>=0 && dx<3 && dy>=0 && dy<3){
                int *newState = st[rear];
                memcpy(newState, s, sizeof(int)*9);
                newState[j] = s[pos];
                newState[pos] = 0;
                if(try_to_insert(rear)){
                    father[rear] = front;  path[rear] = i;
                    rear++;
                }
            }
        } 
        front++;
    }
    ans = rear-1;
}

void print_path(int cur){
    if(cur!=0){
        print_path(father[cur]);
        printf("%c", path_dir[path[cur]]);
    }
}

int main(){
    int T, cas=1;
    scanf("%d", &T);
    while(T--){
        for(int i=0; i<9; ++i){
            scanf("%d", &state[i]);
        }
        bfs();
        printf("Puzzle #%d\n", cas++);
        for(int i=0; i<3; ++i){
            printf("%d %d %d\n", st[ans][i*3], st[ans][i*3+1], st[ans][i*3+2]);
        }
        print_path(ans);
        printf("\n\n");
    }
}

——  生命的意义，在于赋予它意义。
 
  
   
  
  
  

    
   
   
             
   
  
  
  
   
  
  
  

    
   
   
        原创 
    
   
   
   http://blog.csdn.net/shuangde800
    
   
   
    ， By   D_Double  (转载请标明)