1、http://codeforces.com/problemset/problem/400/B
2、题目大意：
有一个n*m的矩阵，其中*代表该格是空的，G代表该格有一个小矮人，S代表该格有一个糖果
每次都要选中所有有小矮人还没到达糖果格的那行，选中的这些行的小矮人都要向右走，如果有一行碰到这两种情况之一，所有行的小矮人都要停止向右运动，两种情况是：小矮人遇到糖果就停止运动，小矮人走到最右边了也停止运动。
求最少执行多少次才能让所有的小矮人都能到达糖果格子，如果不能完成输出-1
3、题目：
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 can 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.
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".
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.
3 4 *G*S G**S *G*S
2
1 3 S*G
-1
4、AC代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 1005
int map[N][N];
int b[N];
int cmp(int a,int b)
{return a<b;
}
int main()
{int n,m;while(scanf("%d%d",&n,&m)!=EOF){//getchar();int flag=0;for(int i=1; i<=n; i++){int p=0,q=0;getchar();for(int j=1; j<=m; j++){scanf("%c",&map[i][j]);if(map[i][j]=='G'){p=j;}else if(map[i][j]=='S'){q=j;}}if(p>q)flag=1;b[i]=q-p;}if(flag==1)printf("-1\n");else{int sum=0;sort(b+1,b+n+1,cmp);for(int i=1; i<=n; i++){if(b[i]!=0){int tmp=b[i];for(int j=i; j<=n; j++){b[j]-=tmp;}sum++;}}printf("%d\n",sum);}}return 0;
}