Forest

You’re a mikemon breeder currently in the middle of your journey to become a mikemon master. Your current obstacle is go through the infamous Biridian Forest.

The forest

The Biridian Forest is a two-dimensional grid consisting of r r r rows and c c c columns. Each cell in Biridian Forest may contain a tree, or may be vacant. A vacant cell may be occupied by zero or more mikemon breeders (there may also be breeders other than you in the forest). Mikemon breeders (including you) cannot enter cells with trees. One of the cells is designated as the exit cell.

The initial grid, including your initial position, the exit cell, and the initial positions of all other breeders, will be given to you. Here’s an example of such grid (from the first example):

0^0

```cpp
#include
using namespace std;
const int N=1007;
char a[N][N];
int n,m,dis[N][N],sx,sy,tx,ty,ans;
const int x_[4]={1,-1,0,0};
const int y_[4]={0,0,1,-1};
struct node{
	int x,y,d;
};
queue<node>q;
void bfs(int tx,int ty){
	q.push(node{tx,ty,0});
	dis[tx][ty]=-1;
	int x,y;
	while(!q.empty()){
		node now=q.front();q.pop();
		for(int i=0;i<4;i++){
			x=now.x+x_[i];
			y=now.y+y_[i];
			dis[x][y]=now.d+1;
		}
	}
}
int main(){
	cin>>n>>m;
	for(int i=1;i<=n;i++)scanf("%s",a[i]+1);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++){
		if(a[i][j]=='S'){sx=i;sy=j;}
		if(a[i][j]=='E'){tx=i;ty=j;}
	}
	bfs(tx,ty);
	for(int i=1;i<=n;i++)
	for(int j=1;j<=m;j++)
		if(dis[i][j]!=0&&dis[i][j]<=dis[sx][sy]&&a[i][j]>='0'&&a[i][j]<='9')ans+=a[i][j]-'0';
	cout<<ans;