HDU 1043 Eight (BFS·八数码·康托展开)

题意  输出八数码问题从给定状态到12345678x的路径

用康托展开将排列相应为整数  即这个排列在全部排列中的字典序  然后就是基础的BFS了

#include <bits/stdc++.h>
using namespace std;
const int N = 5e5, M = 9;
int x[4] = { -1, 1, 0, 0};
int y[4] = {0, 0, -1, 1};
int fac[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320};
int puz[N][M], nex[N], dir[N], vis[N], q[N]; int getCantor(int a[]) //康托展开 将排列转化为整数
{
int ret = 0;
for(int i = 0; i < M; ++i)
{
for(int j = i + 1; j < M; ++j)
if(a[j] < a[i]) ret += fac[M - i - 1];
}
return ret;
} void bfs()
{
int t[M] = {1, 2, 3, 4, 5, 6, 7, 8, 0};
int id = getCantor(t);
dir[id] = -1; memcpy(puz[id], t, sizeof(t));
memset(vis, 0, sizeof(vis)); int r, c, k, nr, nc, nk, nid;
int front = 0, rear = 0;
q[rear++] = id;
vis[id] = 1; while(front < rear)
{
int id = q[front++];
memcpy(t, puz[id], sizeof(t));
for(k = 0; t[k]; ++k); //找0的位置
r = k / 3, c = k % 3; //一维转二维 for(int i = 0; i < 4; ++i)
{
nr = r + x[i], nc = c + y[i], nk = nr * 3 + nc; if(nr < 0 || nr > 2 || nc < 0 || nc > 2) continue;
swap(t[k], t[nk]);
nid = getCantor(t);
memcpy(puz[nid], t, sizeof(t));
swap(t[k], t[nk]); if(vis[nid]) continue;
vis[nid] = 1;
q[rear++] = nid;
nex[nid] = id;
dir[nid] = i;
}
}
} int main()
{
char t[5], sdir[] = "durl";
int s[M], id;
bfs(); while(~scanf("%s", t))
{
s[0] = t[0] == 'x' ? 0 : t[0] - '0';
for(int i = 1; i < M; ++i)
{
scanf("%s", t);
s[i] = t[0] == 'x' ? 0 : t[0] - '0';
} id = getCantor(s);
if(!vis[id]) puts("unsolvable");
else
{
while(dir[id] >= 0)
{
printf("%c", sdir[dir[id]]);
id = nex[id];
}
puts("");
}
}
return 0;
}
//Last modified : 2015-07-05 11:15

Eight

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
 

上一篇:python之读取配置文件模块configparser(三)高级使用---非标准配置文件解析


下一篇:Android判断Service是否运行