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]);

    }

}

巴特西

DFS、栈、双向队列：CF264A- Escape from Stones

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