S1_B_Inna and New Matrix of Candies

Inna and New Matrix of Candies

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Inna likes sweets and a game called the “Candy Matrix”. Today, she came up with the new game “Candy Matrix 2: Reload”.

The field for the new game is a rectangle table of size n × m. Each line of the table contains one cell with a dwarf figurine, one cell with a candy, the other cells of the line are empty. The game lasts for several moves. During each move the player should choose all lines of the matrix where dwarf is not on the cell with candy and shout “Let’s go!”. After that, all the dwarves from the chosen lines start to simultaneously move to the right. During each second, each dwarf goes to the adjacent cell that is located to the right of its current cell. The movement continues until one of the following events occurs:

• some dwarf in one of the chosen lines is located in the rightmost cell of his row;
• some dwarf in the chosen lines is located in the cell with the candy.

The point of the game is to transport all the dwarves to the candy cells.

Inna is fabulous, as she came up with such an interesting game. But what about you? Your task is to play this game optimally well. Specifically, you should say by the given game field what minimum number of moves the player needs to reach the goal of the game.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 1000; 2 ≤ m ≤ 1000).

Next n lines each contain m characters — the game field for the “Candy Martix 2: Reload”. Character “*” represents an empty cell of the field, character “G” represents a dwarf and character “S” represents a candy. The matrix doesn’t contain other characters. It is guaranteed that each line contains exactly one character “G” and one character “S”.

Output

In a single line print a single integer — either the minimum number of moves needed to achieve the aim of the game, or -1, if the aim cannot be achieved on the given game field.

Examples

input

Copy

3 4
*G*S
G**S
*G*S

output

Copy

2

input

Copy

1 3
S*G

output

Copy

-1

题目大意

这道题题目大意才是重点……记一次被题意坑的题

给你一个矩形，每一行都会有个G和S，其余为*代表空。每个回合，G都会向右移动（只能向右移动），当存在有一个G到达S时，回合结束，当所有的G到达S时，游戏结束。问最后最小回合数是多少。如果游戏无法结束，就输出-1。

题目分析

虽然，题目里提到两种回合结束的条件，可是一旦条件一（一个G到达最右边的格子）达成，其实也就说明了这个游戏是无法结束的，因为G只能向右移动。

所以每一行只需要检测S是否在G的左边，如果在左边，就不用继续检测下去了。

然后统计一波每一行G到S的距离。若是相同的距离，则说明在同一个回合内就能到达。这里我图方便就用了集合，开桶记录也是可以的。

参考代码

#include <cstdio>
#include <algorithm>
#include <set>
using namespace std;
int main(int argc, char const *argv[]) {
  int n, m;
  char mp[1005][1005];
  set<int> se;
  scanf("%d%d", &n, &m);
  for(int i = 0; i < n; i++){
    scanf("%s", mp[i]);
  }
  bool flag = 0;
  int min_n = 0x3f3f3f3f, lim = 0x3f3f3f3f, max_n = -1;
  for(int i = 0; i < n; i++){
    int s = 0, g = 0;
    for(int j = 0; j < m; j++){
      if(mp[i][j] == 'S')
        s = j;
      if(mp[i][j] == 'G')
        g = j;
    }
    if(s < g){
      flag = 1;
      break;
    }else{
      se.insert(s - g);
    }
  }
  if(flag)
    printf("%d\n", -1);
  else
    printf("%d\n", se.size());
  return 0;
}