HDU 2102 A计划

【题意】中文题面。

【分析&解题思路】简单BFS。

【AC代码】

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
struct node{
    int x,y,z,time;
}now,nex;
int dir[4][2]={{0,1},{0,-1},{-1,0},{1,0}};
int n,m,t;
char maze[2][11][11];
bool vis[2][11][11];
int bfs();
int main(){
    int tt;
    cin>>tt;
    while(tt--){
        scanf("%d%d%d",&n,&m,&t);
        memset(vis,false,sizeof(vis));
        for(int k=0; k<2; k++){
            for(int i=0; i<n; i++){
                for(int j=0; j<m; j++){
                    cin>>maze[k][i][j];
                }
            }
        }
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if((maze[0][i][j]=='#'&&maze[1][i][j]=='#')||(maze[0][i][j]=='#'&&maze[1][i][j]=='*')|(maze[0][i][j]=='*'&&maze[1][i][j]=='#')){
                    maze[0][i][j]=maze[1][i][j]='*';
                }
            }
        }
        int ans = bfs();
        if(ans==-1) puts("NO");
        else
            puts("YES");
    }
    return 0;
}
bool check(node x){
    if(x.z>=0&&x.z<2&&x.x>=0&&x.x<n&&x.y>=0&&x.y<m&&maze[x.z][x.x][x.y]!='*') return true;
    return false;
}
int bfs(){
    queue<node>qu;
    now.x=0,now.y=0,now.z=0;
    now.time=0;
    vis[0][0][0]=true;
    qu.push(now);
    while(!qu.empty()){
        now=qu.front();
        qu.pop();
        if(now.time>t) return -1;
        if(maze[now.z][now.x][now.y]=='#'){
            if(!vis[now.z^1][now.x][now.y]){
                now.z^=1;
                vis[now.z][now.x][now.y]=true;
            }
        }
        if(maze[now.z][now.x][now.y]=='P') return now.time;
        for(int i=0; i<4; i++){
            nex.z=now.z;
            nex.x=now.x+dir[i][0];
            nex.y=now.y+dir[i][1];
            nex.time=now.time+1;
            if(check(nex)&&!vis[nex.z][nex.x][nex.y]){
                vis[nex.z][nex.x][nex.y]=true;
                qu.push(nex);
            }
        }
    }
    return -1;
}