陷阱題....
把所有的人都能夠比主角早到出口的人全部塞到出口等他就好了
不用想到半路攔截，因為半路攔截也可以跟主角一起到出口在打架
//====================================================================||
// Name : Biridian Forest.cpp ||
// Date : 2013/7/21 下午2:56:36 ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
typedef pair<int,int\> PII;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 1001
#define PB push\_back
#define oo (1<<29)
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
int ny,nx;
int mz\[M\]\[M\];
bool v\[M\]\[M\];
int d\[M\]\[M\];
int ex,ey,bx,by;
int dx\[4\]={0,0,1,-1};
int dy\[4\]={1,-1,0,0};
void solve(){
fill(&d\[0\]\[0\],&d\[M-1\]\[M-1\],oo);
d\[ex\]\[ey\]=0;
queue<PII> q;
q.push(PII(ex,ey));
while(!q.empty()){
int x=q.front().first;
int y=q.front().second;
q.pop();
for(int i=0;i<4;i++){
int nex=x+dx\[i\],ney=y+dy\[i\];
if(v\[nex\]\[ney\])continue;
if(nex<0||nex>=nx||ney<0||ney>=ny)continue;
if(d\[x\]\[y\]+1<d\[nex\]\[ney\]){
d\[nex\]\[ney\]=d\[x\]\[y\]+1;
q.push(PII(nex,ney));
}
}
}
int ans=0;
// debug("%d %d\\n",d\[ex\]\[ey\],d\[2\]\[0\]);
for(int i=0;i<ny;i++){
for(int j=0;j<nx;j++){
if(d\[j\]\[i\]<=d\[bx\]\[by\]){
ans+=mz\[j\]\[i\];
}
}
}
printf("%d\\n",ans);
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d%d%\*c",&ny,&nx)){
char c;
Set(v,0);
Set(mz,0);
for(int i=0;i<ny;i++){
for(int j=0;j<nx;j++){
scanf("%c",&c);
if(c=='E'){
ex=j;ey=i;
}else if(c=='S'){
bx=j;by=i;
}else if(c=='T'){
v\[j\]\[i\]=1;
}else{
mz\[j\]\[i\]=c-'0';
}
}getchar();
}
solve();
}
}