【codeforces 779D】String Game

【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
", ans);
    return 0;
}
原文地址:https://www.cnblogs.com/AWCXV/p/7626587.html