Briefly meaning of the questions:
Now river have \ (n + 2 \) stones, ID \ (0 \) to the \ (n-+. 1 \) , \ (. 1 \) ~ \ (n-\) stones each block has a direction, if is \ (L \) then, the stone then only the frog to jump to the left, if it is \ (R & lt \) only to the right, of course, the \ (0 \) direction of the block of stone is \ (R & lt \ ) .
Now the frog from \ (0 \) jumps \ (the n-+ 1 \) , he should ask how to make him jump jump jump during the longest minimum distance of it? The output of this distance.
Obviously, we consider greedy.
C title is not said than done
You think if you can either jump \ (L \) point, they can jump \ (R \) point, which would you choose it?
Of course \ (R \) point of it!
How to explain it? (Rigorous point)
If \ (L \) in \ (R \) to the left, then jump \ (L \) after, can only have been left until a jump \ (R \) up (can not jump back). Well, you jump at this number \ (L \) in the first \ (L \) time, given that they jump \ (L \) , then you directly choose that \ (R \) is not on the list, but also waste so much step.
If \ (L \) in \ (R \) to the right, then jump \ (L \) After that, you can only return to the \ (R \) point; otherwise you can only have jumped back until another \ (R \) ; then you skip to the path again you jump from the back, I had to jump back, then what is the point, but also a waste of so many steps.
Therefore, the above description has been that we do not jump \ (L \) , just jump \ (R \) is optimal. (Minimum number of steps)
So, the minimum distance of the longest jump much is it? Not that the two adjacent \ (R \) maximum distance of it?
Analyze so much, finally found the nature of the topic, you can directly solve the problem!
Time complexity: \ (O (T \ n-Times) \) .
Because no longer than \ (10 ^ 5 \) , it is \ (O (10 ^ 5) \) .
Space complexity: \ (O (n-) \) .
Actual score: \ (100 pts \) .
#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
const int N=2e5+1;
inline int read(){char ch=getchar();int f=1;while(ch<'0' || ch>'9') {if(ch=='-') f=-f; ch=getchar();}
int x=0;while(ch>='0' && ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();return x*f;}
int T,n;
char s[N];
int main(){
T=read(); while(T--) {
scanf("%s",s);
n=strlen(s);
int k=-1,maxi=0; //k是上一个 R 的位置。因为 s 数组从 0 开始，所以这里从 -1 开始
for(int i=0;i<n;i++)
if(s[i]=='R') {
maxi=max(i-k,maxi); //与上一个 R 的位置的距离
k=i; //更新 R 的位置
}
printf("%d\n",max(maxi,n-k)); //最后一个 R 到终点的距离也算一个答案
}
return 0;
}