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

目录

    • Codeforces Round #704 (Div. 2)-C. Maximum width
      • Problem Description
      • Input
      • Output
      • Sample Input
      • Sample Onput
      • Note
  • 题目大意
  • 题目分析
  • 解题思路
  • AC代码
  • 注意事项

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 s s of length n n n and t t t of length m m m.

A sequence p 1 , p 2 , … , p m p_1, p_2, \ldots, p_m p1,p2,,pm, where 1 ≤ p 1 < p 2 < … < p m ≤ n 1 \leq p_1 < p_2 < \ldots < p_m \leq n 1p1<p2<<pmn, is called beautiful, if s p i = t i s_{p_i} = t_i spi=ti for all i i i from 1 1 1 to m m m. The width of a sequence is defined as max ⁡ 1 ≤ i < m ( p i + 1 − p i ) \max\limits_{1 \le i < m} \left(p_{i + 1} - p_i\right) 1i<mmax(pi+1pi).

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

Input

The first input line contains two integers n n n and m m m ( 2 ≤ m ≤ n ≤ 2 ⋅ 1 0 5 2 \leq m \leq n \leq 2 \cdot 10^5 2mn2105) — the lengths of the strings s s s and t t t.

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

The last line contains a single string t t t of length m m 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 3 3: they are { 1 , 2 , 5 } \{1, 2, 5\} { 1,2,5} and { 1 , 4 , 5 } \{1, 4, 5\} { 1,4,5}.

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

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

In the fourth example there is exactly one beautiful sequence — it is { 1 , 2 } \{1, 2\} { 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aba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bc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 
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。

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