zoukankan      html  css  js  c++  java
  • Codeforces_779_D.String Game_(二分)

    D. String Game
    time limit per test
    2 seconds
    memory limit per test
    512 megabytes
    input
    standard input
    output
    standard output

    Little Nastya has a hobby, she likes to remove some letters from word, to obtain another word. But it turns out to be pretty hard for her, because she is too young. Therefore, her brother Sergey always helps her.

    Sergey gives Nastya the word t and wants to get the word p out of it. Nastya removes letters in a certain order (one after another, in this order strictly), which is specified by permutation of letters' indices of the word ta1... a|t|. We denote the length of word x as |x|. Note that after removing one letter, the indices of other letters don't change. For example, if t = "nastya" and a = [4, 1, 5, 3, 2, 6] then removals make the following sequence of words "nastya"  "nastya"  "nastya"  "nastya"  "nastya"  "nastya"  "nastya".

    Sergey knows this permutation. His goal is to stop his sister at some point and continue removing by himself to get the word p. Since Nastya likes this activity, Sergey wants to stop her as late as possible. Your task is to determine, how many letters Nastya can remove before she will be stopped by Sergey.

    It is guaranteed that the word p can be obtained by removing the letters from word t.

    Input

    The first and second lines of the input contain the words t and p, respectively. Words are composed of lowercase letters of the Latin alphabet (1 ≤ |p| < |t| ≤ 200 000). It is guaranteed that the word p can be obtained by removing the letters from word t.

    Next line contains a permutation a1, a2, ..., a|t| of letter indices that specifies the order in which Nastya removes letters of t (1 ≤ ai ≤ |t|, all ai are distinct).

    Output

    Print a single integer number, the maximum number of letters that Nastya can remove.

    Examples
    input
    ababcba
    abb
    5 3 4 1 7 6 2
    output
    3
    input
    bbbabb
    bb
    1 6 3 4 2 5
    output
    4
    Note

    In the first sample test sequence of removing made by Nastya looks like this:

    "ababcba"  "ababcba"  "ababcba"  "ababcba"

    Nastya can not continue, because it is impossible to get word "abb" from word "ababcba".

    So, Nastya will remove only three letters.

    题意:给定两个字符串str1和str2,再给定一个strlen(str1)的整数序列,整数i表示从str1中删除位置处于i上的字符(即一次删除操作),删除后下标不变。问最多执行多少次删除操作,使剩下的字符串依然可以通过删除字符变成str2.

    妈蛋啊,比赛时弄死没想到啊。。。注意力全放在了两个字符串上。。。结果序列才是关键啊。。。

    思路:二分整数序列,然后找删除后是否合法。。。就这么简单。。。继续努力。。。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<vector>
    #include<algorithm>
    #include<stdlib.h>
    using namespace std;
    #define N 200005
    
    char str1[N],str2[N];
    int num[N];
    bool del[N];
    
    int main()
    {
        scanf("%s%s",str1+1,str2+1);
        int len1=strlen(str1+1),len2=strlen(str2+1);
        for(int i=1;i<=len1;i++)
            scanf("%d",&num[i]);
        int l=1,r=len1,res=0;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            memset(del,0,sizeof(del));
            for(int i=1;i<=mid;i++)
                del[num[i]]=1;
            int pos=1;
            for(int i=1;i<=len1;i++)
            {
                if(del[i]==1)
                    continue;
                if(pos>len2)
                    break;
                if(str1[i]==str2[pos])
                    pos++;
            }
            if(pos>len2)
            {
                res=mid;
                l=mid+1;
            }
            else
                r=mid-1;
        }
        printf("%d
    ",res);
        return 0;
    }
  • 相关阅读:
    elementUI的table分页多选,记住上一页并勾选中,:row-key的使用方法
    如何在vue中使用svg
    父子组件传值,子组件接收不到值,并且无法动态更改video的视频源进行视频播放
    vue项目中如何使用dataform向后台传值
    'eslint'不是内部或外部命令,也不是可运行的程序
    小程序点击分享open-type="share"触发父元素怎么解决?
    vue项目启动报错Module build failed: Error: No PostCSS Config found in:
    eslint在webstorm中有错误警告
    微信小程序 image图片组件实现宽度固定 高度自适应
    JAVA设计模式学习--代理模式
  • 原文地址:https://www.cnblogs.com/jasonlixuetao/p/6451832.html
Copyright © 2011-2022 走看看