779D. String Game 二分 水

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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。

/** @Date    : 2017-05-07 20:26:33
  * @FileName: 779D 二分答案.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoundEarlf@gmail.com)
  * @Link    : https://github.com/Lweleth
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;int p[2*N];
string a, b;
int vis[2*N];int check(int x)
{
int len = a.length();
for(int i = 0; i < len; i++) vis[i] = 0;
for(int i = 0; i < x; i++) vis[p[i] - 1] = 1;int blen = b.length();
int cnt = 0;
for(int i = 0; i < len; i++)
{
if(vis[i])
continue;
if(a[i] == b[cnt])
cnt++;
if(cnt == blen)
return 1;
}
return 0;
}int main()
{
while(cin >> a >> b)
    {
    int r = a.length();
    int l = 1;
    for(int i = 0; i < r; i++)
    scanf("%d", p + i);
    while(l < r)
    {
    int mid = (l + r) >> 1;
    //cout << l << "~" << r << "~" << check(mid) << endl;
    if(check(mid))
    l = mid + 1;
    else r = mid;
    }
    printf("%d\n", l - 1);
    }
    return 0;
}