当前位置：网站首页>Codeforces 814 C. An impassioned circulation of affection （dp）

Codeforces 814 C. An impassioned circulation of affection （dp）

2022-08-10 10:21:00 【51CTO】

Description

Nadeko’s birthday is approaching! As she decorated the room for the party, a long garland of Dianthus-shaped paper pieces was placed on a prominent part of the wall. Brother Koyomi will like it!
Still unsatisfied with the garland, Nadeko decided to polish it again. The garland has n pieces numbered from 1 to n from left to right, and the i-th piece has a colour si, denoted by a lowercase English letter. Nadeko will repaint at most m of the pieces to give each of them an arbitrary new colour (still denoted by a lowercase English letter). After this work, she finds out all subsegments of the garland containing pieces of only colour c — Brother Koyomi’s favourite one, and takes the length of the longest among them to be the Koyomity of the garland.
For instance, let’s say the garland is represented by “kooomo”, and Brother Koyomi’s favourite colour is “o”. Among all subsegments containing pieces of “o” only, “ooo” is the longest, with a length of 3. Thus the Koyomity of this garland equals 3.
But problem arises as Nadeko is unsure about Brother Koyomi’s favourite colour, and has swaying ideas on the amount of work to do. She has q plans on this, each of which can be expressed as a pair of an integer mi and a lowercase letter ci, meanings of which are explained above. You are to find out the maximum Koyomity achievable after repainting the garland according to each plan.

Input

The first line of input contains a positive integer n (1 ≤ n ≤ 1 500) — the length of the garland.
The second line contains n lowercase English letters s1s2… sn as a string — the initial colours of paper pieces on the garland.
The third line contains a positive integer q (1 ≤ q ≤ 200 000) — the number of plans Nadeko has.
The next q lines describe one plan each: the i-th among them contains an integer mi (1 ≤ mi ≤ n) — the maximum amount of pieces to repaint, followed by a space, then by a lowercase English letter ci — Koyomi’s possible favourite colour.

Output

Output q lines: for each work plan, output one line containing an integer — the largest Koyomity achievable after repainting the garland according to it.

Examples input

Examples output

题意

给定一个字符串，有 q q

思路

dp[i][j]dp[i][j] 代表 i i 这一个字符在可以替换 jj

然后我们便可以在 O(n2) O ( n 2 )

当然这道题尺取法也可行，时间复杂度 O(n×q) O ( n × q )

AC 代码

       
       #include <iostream>
       
       #include <cstdio>
       
       #include <cmath>
       
       #include <cstdlib>
       
       #include <cstring>
       
       #include <algorithm>
       
       #define IO ios::sync_with_stdio(false);\
       
       cin.tie(0);\
       
       cout.tie(0);
       
       using 
       
       namespace 
       
       std;
       
       const 
       
       int 
       
       maxn 
       
       = 
       
       1e5
       
       +
       
       10;
       
       const 
       
       int 
       
       mod 
       
       = 
       
       1e9
       
       +
       
       7;
       
       typedef 
       
       long 
       
       long 
       
       LL;
       
       char 
       
       str[
       
       maxn];
       
       int 
       
       dp[
       
       26][
       
       maxn],
       
       n;
       
       void 
       
       init()
       
{
       
       for(
       
       int 
       
       ch
       
       =
       
       0; 
       
       ch
       
       <
       
       26; 
       
       ch
       
       ++)
       
    {
       
       for(
       
       int 
       
       i
       
       =
       
       0; 
       
       i
       
       <
       
       n; 
       
       i
       
       ++)
       
        {
       
       int 
       
       now 
       
       = 
       
       0;
       
       for(
       
       int 
       
       j
       
       =
       
       i; 
       
       j
       
       <
       
       n; 
       
       j
       
       ++)
       
            {
       
       if(
       
       str[
       
       j]
       
       !=
       
       'a'
       
       +
       
       ch)
       
       now
       
       ++;
       
       dp[
       
       ch][
       
       now] 
       
       = 
       
       max(
       
       dp[
       
       ch][
       
       now],
       
       j
       
       -
       
       i
       
       +
       
       1);
       
            }
       
        }
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
       dp[
       
       ch][
       
       i] 
       
       = 
       
       max(
       
       dp[
       
       ch][
       
       i],
       
       dp[
       
       ch][
       
       i
       
       -
       
       1]);
       
    }
       
}
       
       int 
       
       main()
       
{
       
       int 
       
       q;
       
       cin
       
       >>
       
       n
       
       >>
       
       str
       
       >>
       
       q;
       
       init();
       
       while(
       
       q
       
       --)
       
    {
       
       int 
       
       m;
       
       char 
       
       c;
       
       cin
       
       >>
       
       m
       
       >>
       
       c;
       
       cout
       
       <<
       
       dp[
       
       c
       
       -
       
       'a'][
       
       m]
       
       <<
       
       endl;
       
    }
       
       return 
       
       0;
       
}
      
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.