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

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

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

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 $m$ ($2\le m\le n\le 2\cdot 10^5$) — 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

3

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 $\{1,2\}$.

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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。

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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。

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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。

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AC代码

#include 
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。