[题解]八数码问题

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原题链

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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;
}

[题解]八数码问题

标签：else   ret   inline   tmp   pre   map   front   ble   源代码   

原文地址：https://www.cnblogs.com/czy--blog/p/11741849.html