[BFS]正向BFS+hash解决八码问题

八数码问题,可以用单向广搜、双向广搜、A*、IDA等多种方法求解。具体可以参考:八数码的八境界

Description

1
2
3
4
5
6
7
 1  2  3  4 

5 6 7 8

9 10 11 12

13 14 15 x
1
2
3
4
5
6
7
8
9
 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->

Input

1
2
3
4
5
1  2  3 

x 4 6

7 5 8
1
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.

Sample Input

1
2  3  4  1  5  x  7  6  8

Sample Output

1
ullddrurdllurdruldr

题解:

本题一共仅有9!种结果,因此求解方法很多。一开始采用stl进行存储,但一直超时,后来改用hash轻轻松松就过了。判重时9!个排列如果用数组直接保存,每一位保存一个维度,数组开不了那么大。因此可以根据康托展开进行判重,每一种排列对应成一个整形数字,9!种排列一共9!个数字,提高了hash效率。此外,对于x我们暂且当做9处理,而123456789的康托展开是1,因此bfs的终止条件就设为当前状态的康拓展开是否为1。此外,由于本次采用正向bfs,而输出结果时需要输出之前的状态,string储存太慢,queue队列会丢失之前的状态,因此用数组充当队列,用pre追溯上一个状态在队列中的下标。解决代码如下:

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#include<iostream>  
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
const int MAXN = 400000;
int fac[9] = { 1, 1, 2, 6, 24, 120, 720, 5040, 40320};
int dir[4][2] = { {-1,0},{1,0},{0,-1},{0,1} };
int opp[4] = { 'u','d','l','r' };
bool vis[MAXN];
struct node {
int stadus;
int cur[9];
int loc;
char path;
int pre;
};
node qu[MAXN];
int cantor(int s[])
{
int sum = 0;
for (int i = 0; i<9; i++)
{
int num = 0;
for (int j = i + 1; j<9; j++)
if (s[j]<s[i])
num++;
sum += num*fac[8 - i];
}
return sum + 1;
}
int bfs(node now) {
memset(vis, false, sizeof(vis));
int x, y;
int front = 0, end = 0;
node no = now;
qu[end++] = no;
vis[now.stadus] = true;
while (front < end) {
no = qu[front++];
x = no.loc / 3;
y = no.loc % 3;
if (no.stadus == 1)
return front - 1;
for (int i = 0; i < 4; i++) {
node cc = no;
int xx = x + dir[i][0];
int yy = y + dir[i][1];
if (xx < 3 && xx >= 0 && yy < 3 && yy >= 0) {
cc.cur[x * 3 + y] = cc.cur[x * 3 + y] ^ cc.cur[xx * 3 + yy];
cc.cur[xx * 3 + yy] = cc.cur[x * 3 + y] ^ cc.cur[xx * 3 + yy];
cc.cur[x * 3 + y] = cc.cur[x * 3 + y] ^ cc.cur[xx * 3 + yy];
cc.stadus = cantor(cc.cur);
if (!vis[cc.stadus]) {
vis[cc.stadus] = true;
cc.loc = xx * 3 + yy;
cc.path = opp[i];
cc.pre = front - 1;
qu[end++] = cc;
}
}
}
}
return -1;
}
void show(int a) {
if (qu[a].pre) {
show(qu[a].pre);
}
printf("%c", qu[a].path);
}
int main() {
string tmp;
while (getline(cin, tmp)) {
int i = 0, cnt = 0;
node temp;
while (tmp[i]) {
if (tmp[i] == ' ') {
i++;
continue;
}
else if (tmp[i] == 'x') {
temp.loc = cnt;
temp.cur[cnt] = 9;
cnt++;
i++;
}
else {
temp.cur[cnt] = tmp[i] - '0';
cnt++;
i++;
}
}
temp.stadus = cantor(temp.cur);
int ans = bfs(temp);
ans != -1 ? show(ans) : printf("-1");
cout << endl;
}
return 0;
}

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