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]);    }}