[题解]八数码问题

原题链

提交情况

解题思路：

广搜。首先读入，然后特判是不是不需要变换就直接是最后结果（有一个点）。接着入队当前状态，所需步数为\(0\)。然后就是普通广搜的过程。分为\(4\)个方向。每次进行扩展时，都现将表示当前状态的9位数变为一个3×3的矩阵，然后扩展，并判断是否合法。如果合法，则又将\(3×3\)的矩阵变为一个9位数作为函数的返回值。当一遍广搜进行完闭后，就判断是否为目标状态。如果不是，就进行判重。

源代码：

#include <bits/stdc++.h>
using namespace std;
int end_state = 123804765;
map < int,bool > vis;
int a[10][10],x,y;
struct node {
	int state,tm;
};
inline void build(int key){
	if(key % 10 == 0)x = 2,y = 2;a[2][2] = key % 10;key /= 10;
	if(key % 10 == 0)x = 2,y = 1;a[2][1] = key % 10;key /= 10;
	if(key % 10 == 0)x = 2,y = 0;a[2][0] = key % 10;key /= 10;
	if(key % 10 == 0)x = 1,y = 2;a[1][2] = key % 10;key /= 10;
	if(key % 10 == 0)x = 1,y = 1;a[1][1] = key % 10;key /= 10;
	if(key % 10 == 0)x = 1,y = 0;a[1][0] = key % 10;key /= 10;
	if(key % 10 == 0)x = 0,y = 2;a[0][2] = key % 10;key /= 10;
	if(key % 10 == 0)x = 0,y = 1;a[0][1] = key % 10;key /= 10;
	if(key % 10 == 0)x = 0,y = 0;a[0][0] = key % 10;key /= 10;
}
inline int work(int direction,int state) {
	int key = state;
	build(key);
	switch(direction){
		case(1):if(x == 0) return key;swap(a[x][y],a[x - 1][y]);break;//上 
		case(2):if(x == 2) return key;swap(a[x][y],a[x + 1][y]);break;//下
		case(3):if(y == 0) return key;swap(a[x][y],a[x][y - 1]);break;//左
		case(4):if(y == 2) return key;swap(a[x][y],a[x][y + 1]);break;//右 
	} 
	key = 0;
	for(int i = 0;i <= 2;i++){
		for(int j = 0;j <= 2;j++){
			key = key * 10 + a[i][j];
		}	
	}
	return key;
}
queue < node > q;
int main() {
	int now_state;
	cin>>now_state;
	node t;
	t.state = now_state;t.tm = 0;
	q.push(t);
	int tmp;
	if(now_state == end_state)cout<<0,exit(0);
	while(!q.empty()){
		node now,use = q.front();q.pop();
		now_state = work(1,use.state);
		if(now_state == end_state){cout<<use.tm + 1;exit(0);}
		else if(!vis[now_state]){vis[now_state] = true;q.push((node){now_state,use.tm + 1});}
		now_state = work(2,use.state);
		if(now_state == end_state){cout<<use.tm + 1;exit(0);}
		else if(!vis[now_state]){vis[now_state] = true;q.push((node){now_state,use.tm + 1});}
		now_state = work(3,use.state);
		if(now_state == end_state){cout<<use.tm + 1;exit(0);}
		else if(!vis[now_state]){vis[now_state] = true;q.push((node){now_state,use.tm + 1});}
		now_state = work(4,use.state);
		if(now_state == end_state){cout<<use.tm + 1;exit(0);}
		else if(!vis[now_state]){vis[now_state] = true;q.push((node){now_state,use.tm + 1});}
	}
	return 0;
}