neuq 2017: K Multiple Longest Commom Subsequence(LCS变形)

KK has two sequences,A and B, and wants to find the k multiple longest common subsequence.A sequence S is a k multiple common subsequence of A and B if and only if it satisfies the following conditions:
S is a subsequence of A and is a subsequence of B. (A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.)
The length of S is t × k where t is a nonnegative integer. The first element of S is S[1]. If we divide the sequence into t groups with the i- th group containing S[(i − 1) × k + j](1 ≤ j ≤ k), for every element g, it shares the same value with other elements that are in the same group which g belongs to.
For example, [1, 1, 2, 2] is a double common subsequence of [1, 2, 3, 1, 2, 3, 2] and [1, 3, 1, 2, 2]. KK wants to know the maximum length of such sequence.
输入
The first line is an integer T, denoting the number of test cases.
For each test case, the first line are three integers k, n, m, denoting the kind of subsequence, the length of A and the length of B.
The second line are n integers A1 ∼ An, representing the elements of A.
The third line are m integers B1 ∼ Bm, representing the elements of B.
1 ≤ T ≤ 10 , 1 ≤ k, n, m ≤ 103 , 1 ≤ Ai, Bi ≤ 103.
输出
For each test case, output a line with the maximum length of k multiple common subsequence.
样例输入 Copy
3
1 4 5
1 2 3 4
4 1 3 2 4
2 8 7
1 1 2 2 3 3 4 4
1 2 3 1 2 3 3
3 9 9
1 1 1 2 2 2 3 3 3
1 2 3 1 2 3 1 2 3
样例输出 Copy
3
4
3

解析：首先这道题，题目意思就读错了，题目求得是可以分成 t 组，每组保证有 k 个相同的元素，而不是序列长度是k的倍数就可以了。然后在读对题的前提下，当出现相同元素时 dp[i][j]=dp[c[i]-1][d[j]-1]+k,c[i],d[j],分别表示连续出现k个的位置

#include<bits/stdc++.h>
using namespace std;
const int N = 1000+10;
const int M = 500000;
typedef long long LL;
const LL mod = 20180520;
const int maxt=1000000000;
int a[N], b[N], c[N], d[N], dp[N][N];
vector<int>p[N];

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int k, n, m;
        scanf("%d %d %d", &k, &n, &m);
        for(int i=1;i<=n;i++)
            scanf("%d", &a[i]);
        for(int j=1;j<=m;j++)
            scanf("%d", &b[j]);
        memset(dp,0,sizeof(dp));
        memset(c,0,sizeof(c));
        memset(d,0,sizeof(d));

        for(int i=0;i<=1000;i++)p[i].clear();
        for(int i=1;i<=n;i++)
        {
            p[a[i]].push_back(i);
            if(p[a[i]].size()==k)
            {
                c[i]=p[a[i]].front();
                p[a[i]].erase(p[a[i]].begin());
            }
        }

        for(int i=0;i<=1000;i++)p[i].clear();
        for(int i=1;i<=m;i++)
        {
            p[b[i]].push_back(i);
            if(p[b[i]].size()==k)
            {
                d[i]=p[b[i]].front();
                p[b[i]].erase(p[b[i]].begin());
            }
        }

        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
                if(a[i]==b[j]&&c[i]!=0&&d[j]!=0)
                    dp[i][j]=max(dp[i][j],dp[c[i]-1][d[j]-1]+k);
            }
        }
        printf("%d\n",dp[n][m]);
    }

    return 0;
}