codeforces#234_div2_B Inna and New Matrix of Candies
2014-03-07 14:57
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题目地址:cf#234_div2_B
题目大意:给你一个矩阵 每一行有G ,S G代表侏儒,S代表Candy 每次让所有的侏儒往右边走 称为“一步”
直到某一个侏儒碰到最右边或者碰到candy后停下来,
问所有的侏儒碰到candy至少要多少步?
额 其实就是一个水题,,,统计一下差距集合的size就行
代码:
题目大意:给你一个矩阵 每一行有G ,S G代表侏儒,S代表Candy 每次让所有的侏儒往右边走 称为“一步”
直到某一个侏儒碰到最右边或者碰到candy后停下来,
问所有的侏儒碰到candy至少要多少步?
额 其实就是一个水题,,,统计一下差距集合的size就行
代码:
#include<iostream> #include<cstdio> #include<set> using namespace std; char p[1005][1005]; int main() { int n,m; cin>>n>>m; for(int i=0;i<n;i++) scanf("%s",p[i]); bool flag=0; set<int> s; for(int i=0;i<n;i++) { int start,end; for(int j=0;j<m;j++) { if(p[i][j]=='G') start=j; if(p[i][j]=='S') end=j; } if(start<end) s.insert(end-start); else flag=1; } if(flag==0) cout<<s.size()<<endl; else cout<<-1<<endl; }
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