【codeforces 779D】String Game

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【题目链接】:http://codeforces.com/contest/779/problem/D

【题意】

给你一段操作序列;
按顺序依次删掉字符串1中相应位置的字符;
问你最多能按顺序删掉多少个字符;
使得s2是剩下的字符构成的字符串的子列;

【题解】

二分枚举能够按顺序删掉多少个字符m;
然后把1..m相应的字符标记成已经删掉了;
然后O(N)判断s2是不是剩下的字符的子串;
心态炸了.

【完整代码】

#include <bits/stdc++.h>using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%lld",&x)typedef pair<int, int> pii;typedef pair<LL, LL> pll;const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };const double pi = acos(-1.0);const int N = 2e5 + 1000;char s1[N], s2[N];bool bo[N];int a[N], n, l2;bool ok(){    for (int i = 1, j = 1; i <= n && j <= l2; i++)    {        if (!bo[i]) continue;        if (s1[i] == s2[j])        {            j++;            if (j > l2)                return true;        }    }    return false;}int main(){    //freopen("F:\\rush.txt", "r", stdin);    scanf("%s", s1 + 1);    n = strlen(s1 + 1);    scanf("%s", s2 + 1);    l2 = strlen(s2 + 1);    rep1(i, 1, n)        rei(a[i]);    int l = 0, r = n, ans = 0;    while (l <= r)    {        int m = (l + r) >> 1;        rep1(i, 1, n)            bo[a[i]] = true;        rep1(i, 1, m)            bo[a[i]] = false;        if (ok())        {            ans = m;            l = m + 1;        }        else            r = m - 1;    }    printf("%d\n", ans);    return 0;}

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