Igor and his way to work-Codeforce 793B BFS

B. Igor and his way to work

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Woken up by the alarm clock Igor the financial analyst hurried up to the work. He ate his breakfast and sat in his car. Sadly, when he opened his GPS navigator, he found that some of the roads in Bankopolis, the city where he lives, are closed due to road works. Moreover, Igor has some problems with the steering wheel, so he can make no more than two turns on his way to his office in bank.

Bankopolis looks like a grid of n rows andm columns. Igor should find a way from his home to the bank that has no more than two turns and doesn't contain cells with road works, or determine that it is impossible and he should work from home. A turn is a change in movement direction. Igor's car can only move to the left, to the right, upwards and downwards. Initially Igor can choose any direction. Igor is still sleepy, so you should help him.

Input

The first line contains two integers n andm (1 ≤ n, m ≤ 1000) — the number of rows and the number of columns in the grid.

Each of the next n lines contains m characters denoting the corresponding row of the grid. The following characters can occur:

• "." — an empty cell;
• "*" — a cell with road works;
• "S" — the cell where Igor's home is located;
• "T" — the cell where Igor's office is located.

It is guaranteed that "S" and "T" appear exactly once each.

Output

In the only line print "YES" if there is a path between Igor's home and Igor's office with no more than two turns, and "NO" otherwise.

Examples

Input

5 5..S..****.T....****......

Output

YES

Input

5 5S....****.......****..T..

Output

NO

题意：是否能在少于两次转弯次数的条件下从S到达T

分析：搜索题，BFS和DFS都可以，条件不够很容易超时；

我们需要保存到达每个点的需要多少次转弯，可以开一个三维数组vis去存，要注意每次搜索出来不一定是最小的转弯次数；

AC code：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <stack>#include <string>#include <map>using namespace std;#define LL long long#define INF 0x3f3f3f3f#define eps 1const int maxn=100000+50;char map1[1005][1005];int vis[1005][1005][5]; //保存每个点各个方向的最优转弯次数int n,m;int dx[]={0,0,1,-1};int dy[]={1,-1,0,0};struct point{    int x,y;    int dir,turn;       //方向和转弯次数};//是否越界int isOut(int x,int y){    if(x<0||x>=n||y<0||y>=m) return 1;    return 0;}bool bfs(point s){    memset(vis,INF,sizeof vis);    queue<point> q;    q.push(s);    while(!q.empty()){        s=q.front(); q.pop();        if(map1[s.x][s.y]=='T')            return true;        for(int i=0;i<4;i++){            int x=s.x+dx[i];            int y=s.y+dy[i];            int cnt;            if(s.dir==-1||s.dir==i)         //判断是否转弯                cnt=s.turn;            else cnt=s.turn+1;              //转弯次数不能大于2            if(isOut(x,y)||map1[x][y]=='*'||cnt>2) continue;            if(vis[x][y][i]<=cnt) continue; //存在更优解            else{                point t; t.x=x; t.y=y;                t.dir=i; t.turn=cnt;                vis[x][y][i]=cnt;           //更新最优解                q.push(t);            }        }    }    return false;}int main() {    while(~scanf("%d%d",&n,&m)){        for(int i=0;i<n;i++){            scanf("%s",map1[i]);        }        point s;        for(int i=0;i<n;i++){            for(int j=0;j<m;j++){                if(map1[i][j]=='S'){                    s.x=i;                    s.y=j;                    s.turn=0;                    s.dir=-1;                }            }        }        if(bfs(s)) printf("YES\n");        else printf("NO\n");    }}