Description
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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
1
4 4
abab
baba
Example Output
5
题意
对于字符串 S,定义函数 F(S) 为:最少可以将 S 划分为几个连续的子串,使得每个子串仅包含相同的字符。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。你需要将这两个字符串合并成一个长度为 N + M 的字符串 C。C 的每个字符要么来源于 A,要么来源于 B,且来源于 A 的字符的相对顺序应当与在 A 中一致,来源于 B 的字符亦然。
请求出 F(C) 最小可能的值。
思路
unique 以后直接做 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;
}