DFS、栈、双向队列：CF264A- Escape from Stones

题目：

Squirrel Liss liv

Escape from Stonesed in a forest peacefully, but unexpected trouble happens. Stones fall from a mountain. Initially Squirrel Liss occupies an interval [0, 1]. Next, n stones will fall and Liss will escape from
the stones. The stones are numbered from 1 to n in order.

The stones always fall to the center of Liss's interval. When Liss occupies the interval [k - d, k + d] and a stone falls to k, she will escape to the left or to the right. If she escapes to the left, her new interval will be [k - d, k]. If she escapes to the
right, her new interval will be [k, k + d].

You are given a string s of length n. If the i-th character of s is "l" or "r", when the i-th stone falls Liss will escape to the left or to the right, respectively. Find the sequence of stones' numbers from left to right after all the n stones falls.

Input

The input consists of only one line. The only line contains the string s (1 ≤ |s| ≤ 106). Each character in s will be either "l" or "r".

Output

Output n lines — on the i-th line you should print the i-th stone's number from the left.

Example

Input

llrlr

Output

3

5

4

2

1

Input

rrlll

Output

1

2

5

4

3

Input

lrlrr

Output

2

4

5

3

1

解题心得：

1、刚开始看到这个题的时候挺蒙蔽的，第一想法就是使用暴力，但是毫无疑问的是暴力肯定会将精度丢失，即使骗数据过了也容易被Hack。其实这个题有很多的解法

2、先说思路，题意上面说的很明白，当向左的时候最左方的一个是第一个r的前一个，最左方的r的是第一个r出现的时候，可以画一个图了解一下。

3、做法很多，了解了思路之后很简单，直接贴代码：

DFS：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
char a[maxn];
int n;
void dfs(int step)
{
    if(step == n)
        return;
    if(a[step] == 'l')
    {
        dfs(step+1);
        printf("%d ",step+1);
    }
    if(a[step] == 'r')
    {
        printf("%d ",step+1);
        dfs(step+1);
    }
}
int main()
{
    while(scanf("%s",a)!=EOF)
    {
        n = strlen(a);
        dfs(0);
    }
}

栈：

其实和dfs差不多只不过看起来没那么迷糊！

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
char a[maxn];
int main()
{
    scanf("%s",a);
    int len = strlen(a);
    stack <int> s;
    for(int i=0; i<len; i++)
    {
        if(a[i] == 'l')
            s.push(i+1);
        else
            printf("%d ",i+1);
    }
    while(!s.empty())
    {
        printf("%d ",s.top());
        s.pop();
    }
    return 0;
}

双向队列：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+10;
int k[maxn];
char s[maxn];
int main()
{
    scanf("%s",s);
    {
        int len = strlen(s);
        int start = 0,End = len - 1;
        for(int i=0 ;i<len ;i++)
        {
            if(s[i] == 'l')
            {
                k[start] = i+1;
                start++;
            }
            if(s[i] == 'r')
            {
                k[End] = i+1;
                End--;
            }
        }
        for(int i=len-1;i>=0;i--)
            printf("%d ",k[i]);
    }
}