从小明家和小明位置分别用bfs遍历到每个点的最短距离，然后再遍历所有的钥匙的点，最后我们从钥匙中找到到小明家和小明位置距离之和的最小值即可

#include<bits/stdc++.h>
using namespace std;
const int N = 2010, INF = 0x3f3f3f3f;
int n, m;
int vis[N][N];
int ret[2][N][N];
char a[N][N];
int xx[4] = {1,-1,0,0};
int yy[4] = {0,0,1,-1};
void bfs(int xxx, int yyy, int d){
	queue<pair<int,int> > q;
	q.push(make_pair(xxx,yyy));
	vis[xxx][yyy] = 1;
	ret[d][xxx][yyy] = 0;
	while(!q.empty()){
		int tx = q.front().first;
		int ty = q.front().second;
		q.pop();
		for (int i = 0; i < 4; i ++){
			int x = tx + xx[i];
			int y = ty + yy[i];
			if(x >= 0 && x < n && y >= 0 && y < m && a[x][y] != '#' && !vis[x][y]){
				ret[d][x][y] = ret[d][tx][ty] + 1;
				vis[x][y] = 1;
				q.push(make_pair(x,y));
			}
		}
	}
	
}
int main(){
	queue<pair<int, int> > q;
	cin >> n >> m;
	memset(ret,INF, sizeof ret);
	memset(vis,0, sizeof vis);
	int sx, sy;
	int tx, ty;
	for (int i = 0; i < n; i ++){
		cin >> a[i];
		for (int j = 0; j < m; j ++){
			if(a[i][j] == 'P'){
				q.push(make_pair(i,j));
			}else if(a[i][j] == 'S'){
				sx = i;
				sy = j;
			}else if(a[i][j] == 'T'){
				tx = i;
				ty = j;
			}
			
		}
	}
	bfs(sx, sy, 0);
	memset(vis, 0, sizeof vis);
	bfs(tx,ty, 1);
	int ans = INF;
	while(!q.empty()){
		int x = q.front().first;
		int y = q.front().second;
		q.pop();
		ans = min(ans, ret[0][x][y]+ret[1][x][y]);
	}
	cout << ans;
	return 0;
}