B. 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:
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.
Sample test(s)
Input
3 4*G*SG**S*G*S
Output
Input
1 3S*G
Output
-1开始题意理解错了，以为只要一行有个G到了S就停止，记录经过多少个S。。。额。。英语渣真心无脑脑补。。2333333。。。<pre class="n"><p></p><p>题目意思：给n*m的方格，每行中都且仅有一个方格含S和G，每一步可以使所有的G向右移，只到发生下列情况：</p><p>1、某个G已到了最后一列。</p><p>2、某个G已到了S。求把所有的G已到S需要的最少的步数。 </p><p>解题思路：</p><p>统计S-G差出现的个数。</p><p>因为两个差相等的话，他们会一起到达S。也就不要算额外的步数。</p>
#include<cstdio>#include<iostream>#include<cstring>#include<queue>#include<algorithm>#include<vector>using namespace std;const int N = 1020;int n , m;int a, b;int s[N];char str[N][N];int flag;int ans;int main(){ while( scanf("%d%d", &n, &m)!=EOF ) { int flag = 0; memset( s, 0, sizeof(s) ); ans = 0; for(int i=0; i<n; i++) { scanf("%s", str[i]); for(int j=0; j<m; j++) { if( str[i][j]=='G' ) a = j; if( str[i][j]=='S' ) b = j; } if( a<b ) s[ b-a ]++; else flag=1; } for( int i=0; i<m; i++ ) { if( s[i] ) ans++; } if( flag ) printf("-1\n"); else printf("%d\n", ans); } return 0;}
