当前位置：网站首页>CodeChef STMRRG String Merging (dp)

CodeChef STMRRG String Merging (dp)

2022-08-10 10:53:00 【51CTO】

Description

All submissions for this problem are available.
Read problems statements in Mandarin chinese, Russian and Vietnamese as well.
For a string S, let’s define a function F(S) as the minimum number of blocks consisting of consecutive identical characters in S. In other words, F(S) is equal to 1 plus the number of valid indices i such that Si ≠ Si+1.
You are given two strings A and B with lengths N and M respectively. You should merge these two strings into one string C with length N+M. Specifically, each character of C should come either from A or B; all characters from A should be in the same relative order in C as in A and all characters from B should be in the same relative order in C as in B.
Compute the minimum possible value of F(C).

Input

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains two space-separated integers N and M.
The second line contains a single string A with length N.
The third line contains a single string B with length M.

Output

For each test case, print a single line containing one integer — the minimum possible value of F(C).

Constraints

1 ≤ T ≤ 100
1 ≤ N, M ≤ 5,000
1 ≤ sum of N in all test cases ≤ 5,000
1 ≤ sum of M in all test cases ≤ 5,000
strings A, B consist only of lowercase English letters

Example Input

Example Output

题意

对于字符串 S,定义函数 F(S) 为：At least you can S Divide into several consecutive substrings,Makes each substring contain only the same characters.换句话说,F(S) 等于 1 加上满足 Si ≠ Si+1 的合法下标 i 的数量.
给定两个字符串 A 和 B,长度分别为 N 和 M.You need to combine the two strings into one of length N + M 的字符串 C.C Each character of is either derived from A,要么来源于 B,and derived from A The relative order of the characters should be the same as in A 中一致,来源于 B characters as well.
请求出 F(C) 最小可能的值.

思路

unique Do it directly later LCS 即可.

AC 代码

       
       #include <bits/stdc++.h>
       
       #define IO ios::sync_with_stdio(false);\
       
       cin.tie(0);\
       
       cout.tie(0);
       
       using 
       
       namespace 
       
       std;
       
       const 
       
       int 
       
       maxn 
       
       = 
       
       1e4
       
       +
       
       10;
       
       const 
       
       int 
       
       mod 
       
       = 
       
       1e9
       
       +
       
       7;
       
       typedef 
       
       long 
       
       long 
       
       LL;
       
       int 
       
       n,
       
       m;
       
       char 
       
       s1[
       
       maxn],
       
       s2[
       
       maxn];
       
       int 
       
       dp[
       
       maxn][
       
       maxn];
       
       int 
       
       unique_str(
       
       char 
       
       *
       
       s,
       
       int 
       
       len)
       
{
       
       int 
       
       tot 
       
       = 
       
       0;
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <
       
       len; 
       
       i
       
       ++)
       
       if(
       
       s[
       
       i]
       
       !=
       
       s[
       
       tot])
       
       s[
       
       ++
       
       tot] 
       
       = 
       
       s[
       
       i];
       
       return 
       
       tot
       
       +
       
       1;
       
}
       
       int 
       
       main()
       
{
       
       IO;
       
       int 
       
       T;
       
       cin
       
       >>
       
       T;
       
       while(
       
       T
       
       --)
       
    {
       
       cin
       
       >>
       
       n
       
       >>
       
       m;
       
       cin
       
       >>(
       
       s1
       
       +
       
       1);
       
       cin
       
       >>(
       
       s2
       
       +
       
       1);
       
       n 
       
       = 
       
       unique_str(
       
       s1
       
       +
       
       1,
       
       n);
       
       m 
       
       = 
       
       unique_str(
       
       s2
       
       +
       
       1,
       
       m);
       
       for(
       
       int 
       
       i
       
       =
       
       1; 
       
       i
       
       <=
       
       n; 
       
       i
       
       ++)
       
        {
       
       for(
       
       int 
       
       j
       
       =
       
       1; 
       
       j
       
       <=
       
       m; 
       
       j
       
       ++)
       
            {
       
       if(
       
       s1[
       
       i]
       
       ==
       
       s2[
       
       j])
       
       dp[
       
       i][
       
       j] 
       
       = 
       
       dp[
       
       i
       
       -
       
       1][
       
       j
       
       -
       
       1] 
       
       + 
       
       1;
       
       else
       
       dp[
       
       i][
       
       j] 
       
       = 
       
       max(
       
       dp[
       
       i
       
       -
       
       1][
       
       j],
       
       dp[
       
       i][
       
       j
       
       -
       
       1]);
       
            }
       
        }
       
       cout
       
       <<
       
       n
       
       +
       
       m
       
       -
       
       dp[
       
       n][
       
       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.