Code Forces 645A Amity Assessment-优快云博客

本文介绍了一种判断两个2x2滑动拼图是否可以通过移动达到相同配置的方法。利用广度优先搜索算法，通过定义节点状态并记录已访问状态来避免重复搜索，最终确定两个初始状态的拼图是否能够达成一致。

A. Amity Assessment
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Bessie the cow and her best friend Elsie each received a sliding puzzle on Pi Day. Their puzzles consist of a 2 × 2 grid and three tiles labeled ‘A’, ‘B’, and ‘C’. The three tiles sit on top of the grid, leaving one grid cell empty. To make a move, Bessie or Elsie can slide a tile adjacent to the empty cell into the empty cell as shown below:

In order to determine if they are truly Best Friends For Life (BFFLs), Bessie and Elsie would like to know if there exists a sequence of moves that takes their puzzles to the same configuration (moves can be performed in both puzzles). Two puzzles are considered to be in the same configuration if each tile is on top of the same grid cell in both puzzles. Since the tiles are labeled with letters, rotations and reflections are not allowed.

Input
The first two lines of the input consist of a 2 × 2 grid describing the initial configuration of Bessie’s puzzle. The next two lines contain a 2 × 2 grid describing the initial configuration of Elsie’s puzzle. The positions of the tiles are labeled ‘A’, ‘B’, and ‘C’, while the empty cell is labeled ‘X’. It’s guaranteed that both puzzles contain exactly one tile with each letter and exactly one empty position.

Output
Output “YES”(without quotes) if the puzzles can reach the same configuration (and Bessie and Elsie are truly BFFLs). Otherwise, print “NO” (without quotes).

Examples
input
AB
XC
XB
AC
output
YES
input
AB
XC
AC
BX
output
NO

我直接暴力来的

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <queue>
#include <map>

using namespace std;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int vis[5000];
struct Node
{
    int a[2][2];
    int s;
    int posx;
    int posy;
    Node(){};
    Node(int a[2][2],int s,int posx,int posy)
    {
        this->s=s;
        this->posx=posx;
        this->posy=posy;
        memcpy(this->a,a,sizeof(this->a));
    }
};
Node st,ed;
map<char,int>m;

int kt(int a[2][2])
{
    int num=0;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++)
           num=num*10+a[i][j];
    return num;
}
int bfs(Node st)
{
    queue<Node>q;
    q.push(st);
    while(!q.empty())
    {
        Node term=q.front();
        q.pop();
        if(term.s==ed.s)
        {
            return 1;
        }

        for(int i=0;i<4;i++)
        {
            int xx=term.posx+dir[i][0];
            int yy=term.posy+dir[i][1];
            if(xx<0||xx>=2||yy<0||yy>=2)
                continue;
            swap(term.a[term.posx][term.posy],term.a[xx][yy]);
            int state=kt(term.a);
            if(vis[state])
            {
                swap(term.a[term.posx][term.posy],term.a[xx][yy]);
                continue;
            }
            vis[state]=1;
            q.push(Node(term.a,state,xx,yy));
            swap(term.a[term.posx][term.posy],term.a[xx][yy]);
        }
    }
    return 0;
}
int main()
{
    m['A']=1;m['B']=2;m['C']=3;m['X']=0;
    char b1[2][2];
    memset(vis,0,sizeof(vis));
    for(int i=0;i<=1;i++)
        scanf("%s",b1[i]);
    for(int i=0;i<=1;i++)
        for(int j=0;j<=1;j++)
        {
            st.a[i][j]=m[b1[i][j]];
            if(b1[i][j]=='X')
                st.posx=i,st.posy=j;
        }
    st.s=kt(st.a);
    for(int i=0;i<=1;i++)
        scanf("%s",b1[i]);
    for(int i=0;i<=1;i++)
        for(int j=0;j<=1;j++)
        {
            ed.a[i][j]=m[b1[i][j]];
            if(b1[i][j]=='X')
                ed.posx=i,ed.posy=j;
        }
    ed.s=kt(ed.a);
    if(bfs(st))
        printf("YES\n");
    else
        printf("NO\n");
    return 0;

}