Playf and Astringent Map

The first line contains two integers n and m ( $1\le n,m \le100$ ) — size of the Astringent Map.

Then, n lines follow. Each line contains m characters, represents the Astringent Map.

It is guaranteed that the Astringent Map only contains 'o', '.', 'P' and 'X', where 'P' represents the initial position of Playf and 'X' represents the initial position of Xu. There is exactly one 'X' and one 'P' on the Astringent Map.

C 解法, 执行用时: 3ms, 内存消耗: 524K, 提交时间: 2022-01-02 15:13:42

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int n,m;
int count=0;
char map[100][100];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

int check(int x,int y){
	if(y>=0&&y<m&&x>=0&&x<n&&map[x][y]=='.'||map[x][y]=='P'){
		return 1;
	}return 0;
}

void dfs(int x,int y){
	if(check(x,y)){
		map[x][y]='o';
	for(int i=0;i<4;i++){
		int xx=x+dir[i][0];
		int yy=y+dir[i][1];
			if(map[xx][yy]=='X'){
				count++;
				return ;
			}else{
				dfs(xx,yy);
			}
		}
	}
	return ;
}
int main(void)
{
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		scanf("%s",map[i]);
	}
	int x=0,y=0;
	int flag=0;
	for(y=0;y<m;y++){
		for(x=0;x<n;x++){
			if(map[x][y]=='P'){
				flag++;
				break;
			}
		}
		if(flag)break;
	}
	dfs(x,y);
	if(count)printf("Playftql");
	else printf("Playftcl");
	return 0;
}

C++ 解法, 执行用时: 4ms, 内存消耗: 640K, 提交时间: 2022-01-03 21:30:11

#include<iostream>
using namespace std;
int n,m,q,w;
char a[110][110];
int dx[4]={0,1,-1,0};
int dy[4]={1,0,0,-1};
int nx,ny;
void dfs(int x,int y){
	a[x][y]='o';
	for(int i=0;i<4;i++){
		nx=x+dx[i];ny=y+dy[i];
		if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&a[nx][ny]!='o'){
			dfs(nx,ny);
		}
	}
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			cin>>a[i][j];
			if(a[i][j]=='P'){
				q=i;w=j;
			}		
		}
	}
	dfs(q,w);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			if(a[i][j]=='X'){
				cout<<"Playftcl";
				return 0;
			}
		}
	}
	cout<<"Playftql";
	return 0;
}

详情