BFS maze

Links: https://ac.nowcoder.com/acm/challenge/terminal
Source: Cattle-off network

Xiao Ming is now playing a game, the game came to the teaching points, it is a maze of N * M matrix.

Xiao Ming's starting point on the map with "S" to indicate the end point with "E" to represent an obstacle by "#" is represented by open space "." To represent.

It can not be an obstacle. If Bob is now at a point (x, y), the next step can only go four adjacent grid in one of: (x + 1, y), (x-1, y), (x, y +1), (x, y-1);

Xiao Ming wants to know, he can now come to the end from the beginning.

Enter a description:

This title contains multiple sets of data. 
Each set of data input to two numbers N, M 
Next N lines of M symbols to represent a map of the state. 
Data range: 
2 <= N, M <= 500 
ensure that there is a starting point S, while ensuring an end E.

Output Description:

Each data output line, if Bob can come to the end from the beginning, then the output Yes, otherwise output No

Example 1

Entry

copy

3 3
S..
..E
...
3 3
S##
###
##E

Export

copy

Yes
No 

problem-solving ideas: classical BFD

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}
typedef long long ll;
const int maxn = 1e5+10;
char a[600][600];
int vis[600][600];
struct node{
    int x,y;
};
int d[4][2]={0,1,0,-1,1,0,-1,0};
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        node S;//记录起点
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                cin>>a[i][j];
                if(a[i][j]=='S'){
                    S.x=i;
                    Sy = j;
                }
            }
        }
        int flag=0;
        Memset (VIS, 0 , the sizeof (VIS)); // Clear tag array 
        Queue <Node> Q; // define a queue 
        q.push (S); // inject a starting point 
        the while (! q.empty ()) {
            TEMP Node = q.front (); // Analyzing 
            q.pop (); // pop-up point for the first 
            IF (A [temp.x] [temp.y] == ' E ' ) {
                flag=1;
                break;
            }
            vis[temp.x][temp.y]=1;
            for(int i=0;i<4;i++){
                node R;
                R.x=temp.x+d[i][0];
                R.y=temp.y+d[i][1];
                if(R.x>=1&&R.x<=n&&R.y>=1&&R.y<=m&&vis[R.x][R.y]==0&&a[R.x][R.y]!='#'){
                    manner [R x] [y] = 1 ;
                    q.push (R & lt); // stack idle determination 
                }
            } 
        }
        if(flag==1){
            printf("Yes\n"); 
        } 
        else{
            printf("No\n");
        } 
    }
    return 0;
}

AC code 2;

 

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
char map[600][600];
char s[100];
bool vis[600][600];
struct Node{
    int x,y;
};
int d[4][2]={0,1,0,-1,1,0,-1,0};
int main()
{
    int N,M;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        Node S;
        gets(s);//getchar();
        for(int i=1;i<=N;i++){
            for(int j=1;j<=M;j++){
                scanf("%c",&map[i][j]);
                if(map[i][j]=='S'){
                    S.x=i;
                    Sy = j;
                }
            }
            gets(s);
        }
        int flag=0;
        memset(vis,0,sizeof(vis));
        queue<Node>q;
        q.push(S);
        while(!q.empty())
        {
            Node temp=q.front();
            q.pop();
            if(map[temp.x][temp.y]=='E')
            {flag=1;break;}
            vis[temp.x][temp.y]=1;
            for(int i=0;i<4;i++)
            {
                Node R;
                R.x=temp.x+d[i][0];R.y=temp.y+d[i][1];
                if(R.x<1||R.y<1||R.x>N||R.y>M) continue;
                if(!vis[R.x][R.y]&&map[R.x][R.y]!='#')
                {
                    manner [R x] [y] = 1 ;
                    q.push(R);
                }
            }
        }
        if(flag) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}