CF264A

题目链接

    题目大意是： Squirrel Liss 住在一个平静的深林, 突然不幸发生了. 山上的石头滚下来了. 然后会有n个石头落下来, Squirrel Liss 将会躲避落石. 石头从1到n依次编号. 落石每次都正好落在Squirrel Liss 居住范围的中间(p-k, p+k). 然后他可以选择向左(p-k, p)或向右(p, p+k).我们将会得到一串由lr组成的字符串来指示他向左或向右.最后从左向右输出石头的编号.
    刚开始分析了一下, 发现不就是个双向链表(用数组模拟)插入和输出(O(N)). 样例过了就自信提交. 呵呵time out.  (可能常数项大了)

双向链表(time out)

#include<cstdio>
#include<cstring>
using namespace std;
char str[1000001];
int b[1000001];
int f[1000001];
int main()
{
    int p = 1, s = 1;
    scanf("%s", str);
    memset(b, 0, sizeof(b));
    memset(f, 0, sizeof(f));
    for ( int i = 0; i < strlen(str)-1; i ++ )
    {
        if ( str[i] == 'l' )
        {
            if ( f[p] == 0 )
            {
                b[i+2] = p;
                f[p] = i+2;
                s = i+2;
            }
            else
            {
                int pp;
                pp = f[p];
                f[i+2] = pp;
                b[pp]  = i+2;
                b[i+2] = p;
                f[p] = i+2;
            }
        }
        else
        {
            if ( b[p] == 0 )
            {
                f[i+2] = p;
                b[p] = i+2;
            }
            else
            {
                int pp;
                pp = b[p];
                b[i+2] = pp;
                f[pp]  = i+2;
                f[i+2] = p;
                b[p] = i+2;
            }
        }
        p = i + 2;
    }
    printf("%d", s);
    while ( s = b[s] )  printf("\n%d", s);

    return 0;
}

    每次向左都会导致下一个石头左移, 所以后落得石头在左边(filo), 每次右移会得到之后的石头位置都在右边. 然后...递归一下...

#include<cstdio>
#include<cstring>
using namespace std;
char str[1000001];
void dfs(int n)
{
    if ( str[n] == '\0' )   return;
    if ( str[n] == 'l' )
    {
        dfs(n+1);
        printf("%d\n", n+1);
    }
    else
    {
        printf("%d\n", n+1);
        dfs(n+1);
    }
}

int main()
{
    scanf("%s", str);
    dfs(0);
    return 0;
}

    其实会发现规律, 顺序输出r, 逆序输出l

#include<cstdio>
#include<cstring>
using namespace std;
char str[1000001];
int ans[1000001];
int main()
{
    scanf("%s", str);
    int l = 0;
    int r = strlen(str) - 1;
    for ( int i = 0; i < strlen(str); i ++ )
    {
        if ( str[i] == 'l' )    ans[r --] = i+1;
        else                    ans[l ++] = i+1;
    }
    for ( int i = 0; i < strlen(str); i ++ )    printf("%d\n", ans[i]);
    return 0;
}