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Codeforces Round #704 (Div. 2)-C. Maximum width

传送门 Time Limit: 2 seconds Memory Limit: 512 megabytes

Problem Description

Your classmate, whom you do not like because he is boring, but whom you respect for his intellect, has two strings: $s$ of length $n$ and $t$ of length $m$ .

A sequence $p_1, p_2, \ldots, p_m$ , where $1 \leq p_1 < p_2 < \ldots < p_m \leq n$ , is called beautiful, if $s_{p_i} = t_i$ for all $i$ from $1$ to $m$ . The width of a sequence is defined as $\max\limits_{1 \le i < m} \left(p_{i + 1} - p_i\right)$ .

Please help your classmate to identify the beautiful sequence with the maximum width. Your classmate promised you that for the given strings $s$ and $t$ there is at least one beautiful sequence.

Input

The first input line contains two integers $n$ and $m$ ( $2 \leq m \leq n \leq 2 \cdot 10^5$ ) — the lengths of the strings $s$ and $t$ .

The following line contains a single string $s$ of length $n$ , consisting of lowercase letters of the Latin alphabet.

The last line contains a single string $t$ of length $m$ , consisting of lowercase letters of the Latin alphabet.

It is guaranteed that there is at least one beautiful sequence for the given strings.

Output

Output one integer — the maximum width of a beautiful sequence.

Sample Input

5 3
abbbc
abc

Sample Onput

Note

In the first example there are two beautiful sequences of width $3$ : they are $\{1, 2, 5\}$ and $\{1, 4, 5\}$ .

In the second example the beautiful sequence with the maximum width is $\{1, 5\}$ .

In the third example there is exactly one beautiful sequence — it is $\{1, 2, 3, 4, 5\}$ .

In the fourth example there is exactly one beautiful sequence — it is $\{1, 2\}$ .

题目大意

两个字符串s和t长度分别是n和m。在s从左到右选一些字母组成t。问s中所有选中的字符中，最大间距是多少。

题目保证能从s中选出t。

题目分析

要使间距最大，就要t中的某两个相邻的字母在s中选择时，第一个字母尽可能往前，第二个字母尽可能往后，这样才能使间距最大。

如s(abbbc)中选择t(abc)，看t中相邻两个字母，先看a和b，a要选的尽可能靠前（其实就一种选择），所以选第1个。b要选的尽可能靠后（可以选第2~4个），所以选第4个。这样的话它的最大间距就是4-3=3。同理看b和c时，应分别选择第2个和第5个，间距同样是5-2=3。所以最大间距是3。

解题思路

先从前往后遍历一遍，得到t中每个字母最靠前能选到s的第几个；再从后往前遍历一遍，得到t中每个字母最靠后能选到s的第几个。考虑所有t中相邻的两个字母，看看第一个字母最靠前，第二个字母最靠后，间距是多少。更新最大间距，直到遍历完t。

AC代码

#include <bits/stdc++.h>
using namespace std;
#define mem(a) memset(a, 0, sizeof(a))
#define dbg(x) cout << #x << " = " << x << endl
#define fi(i, l, r) for (int i = l; i < r; i++)
#define cd(a) scanf("%d", &a)
typedef long long ll;
char s[200010];
char t[200010];
int zuiQian[200010]; //最靠前能选到s中的第几个
int zuiHou[200010]; //最靠后能选到s中的第几个
int main()
{
    int n, m, M = 1;
    scanf("%d%d", &n, &m);
    scanf("%s", s);
    scanf("%s", t);
    int lastT = -1;
    for (int i = 0; i < m; i++)
    {
        while (s[++lastT] != t[i]) //找到s中第一个没有选过的并且与t的下一个相同的字母。
            ;
        zuiQian[i] = lastT;
    }
    lastT = n;
    for (int i = m - 1; i >= 0; i--)
    {
        while (s[--lastT] != t[i])
            ;
        zuiHou[i] = lastT;
    }
    for (int i = 1; i < m; i++)
    {
        M = max(M, zuiHou[i] - zuiQian[i - 1]); //更新最大值
    }
    printf("%d\n", M);
    return 0;
}

注意事项

while (s[++lastT] != t[i]);是打比赛时确定s中下一个位置的简便写法。

学会的话打比赛会快亿点点，但是小心不要用错了哦。

同步发文于我的CSDN

Codeforces Round #704 (Div. 2)-C. Maximum width-题解