【Codeforces Round #402 (Div. 1)】Codeforces 778A String Game

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. 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.

首先考虑如何判断一个字符串能否通过去掉一些字符变成另一个。显然只要O(n)贪心地扫描一遍。
这样套上二分答案就可以了，复杂度O(nlogn)。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;char t[200010],p[200010];int flag[200010],a[200010],n,m;bool ok(int x){    int j=1;    for (int i=1;i<=x;i++) flag[a[i]]=0;    for (int i=x+1;i<=n;i++) flag[a[i]]=1;    for (int i=1;i<=n&&j<=m;i++)        if (flag[i]&&t[i]==p[j]) j++;    return j==m+1;}int main(){    int l,r,mid;    scanf("%s%s",t+1,p+1);    n=strlen(t+1);    m=strlen(p+1);    for (int i=1;i<=n;i++) scanf("%d",&a[i]);    l=0,r=n;    while (l<r)    {        mid=(l+r+1)/2;        if (ok(mid)) l=mid;        else r=mid-1;    }    printf("%d\n",l);}