Codeforces 778A（二分答案）

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本文介绍Codeforces778A题目StringGame的解决思路，通过二分查找确定最多可删除字符数，确保剩余字符能构成目标字符串p。文章提供了一种使用优先队列进行字符匹配的有效方法。

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Codeforces 778A String Game

题目：
String Game
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard 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.

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.

题意：
给定两个字符串t和p，以及a数组，按照a数组顺序删除字符串中字符，问最多可以删掉多少字符是剩下的字符串在只删去一些字符而不改变字符位置的情况下可以构成p字符串。

就是裸的二分答案，然而想了好久才想到2333，对总长度二分，然后判断区间内的字符能否组成p字符串，（我是借助一个优先队列，把字符串放在其中判断，使其按照原始顺序，不过似乎有好多种方法）

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
#include <stack>
#include <set>
#include <queue>
#include <vector>
using namespace std;
char s[200005];
char p[200005];
char b[200005];
int a[200005];
struct xx
{
    char x;
    int y;
    bool operator < (xx m) const
    {
        return y>m.y;
    }
    xx(char m,int n){x=m,y=n;}
    xx(){}
};
priority_queue<struct xx> q;
bool check()
{
    int num=0,len=strlen(p);
    while(!q.empty()){
        struct xx x;
        x=q.top();
        q.pop();
        if(x.x==p[num]){
            num++;
        }
        if(num==len) return true;
    }
    return false;

}
int main()
{
    scanf("%s %s",s,p);
    int len=strlen(s);
    for(int i=0;i<len;i++){
        scanf("%d",&a[i]);
    }
    int l=0,r=len;
    while(l<r)
    {
        int mid=(l+r)/2;
        while(!q.empty()) q.pop();
        for(int i=mid;i<len;i++){
            q.push(xx(s[a[i]-1],a[i]));
        }
        if(check()){
            l=mid+1;

        }
        else r=mid;
    }
    printf("%d\n",r-1);
    return 0;
}