Codeforces_779_D.String Game_(二分)
2 seconds
512 megabytes
standard input
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 t: a1... 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.
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).
Print a single integer number, the maximum number of letters that Nastya can remove.
ababcba
abb
5 3 4 1 7 6 2
3
bbbabb
bb
1 6 3 4 2 5
4
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+,str2+);
int len1=strlen(str1+),len2=strlen(str2+);
for(int i=;i<=len1;i++)
scanf("%d",&num[i]);
int l=,r=len1,res=;
while(l<=r)
{
int mid=(l+r)>>;
memset(del,,sizeof(del));
for(int i=;i<=mid;i++)
del[num[i]]=;
int pos=;
for(int i=;i<=len1;i++)
{
if(del[i]==)
continue;
if(pos>len2)
break;
if(str1[i]==str2[pos])
pos++;
}
if(pos>len2)
{
res=mid;
l=mid+;
}
else
r=mid-;
}
printf("%d\n",res);
return ;
}
Codeforces_779_D.String Game_(二分)的更多相关文章
- HDU5853 Jong Hyok and String(二分 + 后缀数组)
题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5853 Description Jong Hyok loves strings. One da ...
- acdream1116 Gao the string!(hash二分 or 后缀数组)
问题套了一个斐波那契数,归根结底就是要求对于所有后缀s[i...n-1],所有前缀在其中出现的总次数.我一开始做的时候想了好久,后来看了别人的解法才恍然大悟.对于一个后缀来说 s[i...n-1]来说 ...
- Codeforces Round #402 D String Game(二分)
[题目类型]二分答案 &题解: 只要你想到二分答案就不是难题了,但我当时确实是想不到. [时间复杂度]\(O(nlogn)\) &代码: #include <cstdio> ...
- Codeforces 799D. String Game 二分
D. String Game time limit per test:2 seconds memory limit per test:512 megabytes input:standard inpu ...
- 779D. String Game 二分 水
Link 题意: 给出两字符串$a$,$b$及一个序列,要求从前往后按照序列删掉$a$上的字符,问最少删多少使$b$串不为a的子串 思路: 限制低,直接二分答案,即二分序列位置,不断check即可. ...
- codeforces D. Mahmoud and Ehab and the binary string(二分)
题目链接:http://codeforces.com/contest/862/submission/30696399 题解:这题一看操作数就知道是二分答案了.然后就是怎么个二分法,有两种思路第一种是找 ...
- Codeforces 862D. Mahmoud and Ehab and the binary string 【二分】(交互)
<题目链接> 题目大意: 有一个长度为n(n<1000)的01串,该串中至少有一个0和一个1,现在由你构造出一些01串,进行询问,然后系统会给出你构造的串与原串的 Hamming ...
- Codeforces.862D.Mahmoud and Ehab and the binary string(交互 二分)
题目链接 \(Description\) 有一个长为\(n\)的二进制串,保证\(01\)都存在.你可以询问不超过\(15\)次,每次询问你给出一个长为\(n\)的二进制串,交互库会返回你的串和目标串 ...
- 【CF676C】Vasya and String(二分查找,线性扫描尺取法)
题意: 给出一个长度为n的字符串,只有字符'a'和'b'.最多能改变k个字符,即把'a'变成'b'或把'b'变成'a'. 问改变后的最长连续相同字符的字串长度为多少. 首先是二分查找,好想也好写 .. ...
随机推荐
- AutoCAD菜单加载失败 找不到文件mnc 怎么办
菜单加载失败,找不到文件 SWFILECONV(mnu/mns/mnc) 找到CAD安装目录下的swfileconv.arx文件,用记事本打开,清空内容,然后保存即可.
- jquery 联动 年月日
<html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>JQuery实例 - 生成年 ...
- 【剑指Offer学习】【面试题65:滑动窗体的最大值】
题目:给定一个数组和滑动窗体的大小,请找出全部滑动窗体里的最大值. 举例说明 比如,假设输入数组{2,3,4,2,6,2,5,1}及滑动窗体的大小.那么一共存在6个滑动窗体,它们的最大值分别为{4,4 ...
- 【Struts2五】ValueStack以及ognl表达式二(经常使用标签)
Ognl经常使用标签: 1.s:debug 假设把该标签放入到s:iterator中能够看到当前正在迭代的元素的状态 2.s:property 1.输出 ...
- SpringMVC 运行过程
加入jar包 在web.xml中配置DispatherServlet 加入SpringMVC配置文件 编写处理请求的处理器,并表示为处理器 编写视图 可见的运行流程: 实际的运行流程:
- android判断正在使用的网络类型 0.不知道网络类型;1、2G;2、3G;3、4g;4、wifi
判断正在使用的网络类型 0.不知道网络类型:1.2G:2.3G:3.4g:4.wifi /** Unknown network class. {@hide} */ public static fina ...
- MySQL:MySQL安装
ylbtech-MySQL:MySQL安装 mysql5.5.27_win64_zol.msi 1.返回顶部 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14 ...
- 词典(一) 跳转表(Skip table)
词典,顾名思义,就是通过关键码来查询的结构.二叉搜索树也可以作为词典,不过各种BST,如AVL树.B-树.红黑树.伸展树,结构和操作比较复杂,而且理论上插入和删除都需要O(logn)的复杂度. 在词典 ...
- PHP面向对象技术(全面讲解)
作者:高洛峰 来源:<PHPer>杂志 1.面向对象的概念 面向对象编程(Object Oriented Programming, OOP, 面向对象程序设计)是一种计算机 ...
- PCB DotNetCore Swagger生成WebAPI文档配置方法
在.net framework框架下可以使用WebApiTestClientWebApi生成WebAPI接口文档与方便接口测试用,而在DotnetCore却没有找到这个工具了,baidu查找一下发现有 ...