AGC007 A-Shik and Stone (simulation)

Title:

solution:

令(x,y)=(1,1),即当前位置.
然后模拟行走,如果(x,y+1)='#',那么将(x,y)修改为(x,y+1),
否则如果(x+1,y)='#',那么将(x,y)修改为(x+1,y).
如果不能走则结束流程.

标记上述走法走过的所有格子,
然后遍历每个格子,如果存在s[i][j]='#'但是却未被标记过,那么无解.

code：

#include <bits/stdc++.h>
#define int long long
#define PI pair<int,int>
using namespace std;
const int maxm=2e3+5;
int mark[maxm][maxm];
char s[maxm][maxm];
int n,m;
void solve(){
    
    
    cin>>n>>m;
    for(int i=1;i<=n;i++){
    
    
        cin>>(s[i]+1);
    }
    if(s[1][1]!='#'){
    
    
        cout<<"Impossible"<<endl;
        return ;
    }
    int x=1,y=1;
    mark[1][1]=1;
    while(1){
    
    
        if(y+1<=m&&s[x][y+1]=='#'){
    
    
            y++;
            mark[x][y]=1;
        }else if(x+1<=n&&s[x+1][y]=='#'){
    
    
            x++;
            mark[x][y]=1;
        }else{
    
    
            break;
        }
    }
    for(int i=1;i<=n;i++){
    
    
        for(int j=1;j<=m;j++){
    
    
            if(s[i][j]=='#'&&!mark[i][j]){
    
    
                cout<<"Impossible"<<endl;
                return ;
            }
        }
    }
    cout<<"Possible"<<endl;
}
signed main(){
    
    
    ios::sync_with_stdio(0);
    solve();
    return 0;
}

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