河北省省赛重现赛-K Multiple Longest Commom Subsequence

2017: K Multiple Longest Commom Subsequence

描述

题目描述：

KK has two sequences, AAA and BBB, and wants to find the kkk multiple longest common subsequence.A sequence SSS is a kkk multiple common subsequence of AAA and BBB if and only if it satisfies the following conditions:

SSS is a subsequence of AAA and is a subsequence of BBB. (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 SSS is t×kt\times kt×k where ttt is a nonnegative integer. The first element of SSS is S[1]S[1]S[1]. If we divide the sequence into ttt groups with the iiith group containing S[(i−1)×k+j](1≤j≤k)S[(i - 1) \times k + j] (1 \leq j \leq k)S[(i−1)×k+j](1≤j≤k), for every element ggg, it shares the same value with other elements that are in the same group which ggg belongs to.

For example, [1,1,2,2][1, 1, 2, 2][1,1,2,2] is a double common subsequence of [1,2,3,1,2,3,2][1, 2, 3, 1, 2, 3, 2][1,2,3,1,2,3,2] and [1,3,1,2,2][1, 3, 1, 2, 2][1,3,1,2,2]. KK wants to know the maximum length of such sequence.

输入：

The first line is an integer TTT, denoting the number of test cases.

For each test case, the first line are three integers kkk , nnn, mmm, denoting the kind of subsequence, the length of AAA and the length of BBB.

The second line are nnn integers A1∼AnA_1 \sim A_nA1∼An, representing the elements of AAA.

The third line are mmm integers B1∼BmB_1 \sim B_mB1∼Bm, representing the elements of BBB.

1≤T≤101 \leq T \leq 101≤T≤10 , 1≤k,n,m≤1031\leq k, n, m \leq 10^31≤k,n,m≤103, 1≤Ai,Bi≤1031\leq A_i, B_i \leq 10^31≤Ai,Bi≤103.

输出：

For each test case, output a line with the maximum length of kkk multiple common subsequence.

样例输入

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

3
4
3

这题最重要的一点就是把 k个相同的数当成一个整体来看

题解：最长公共子序列（LCS），这个问题是要分第a[i]==b[i] 和 a[i]!=b[j]，如果a[i]==b[i] 那么就要看是否有前 k个相同的a[i] ，和前k个相同的b[i]，并分别记录最前面的那个下标为pa[i]，和pb[i]，如果pa[i]！=0 && pb[i]!=0 那么就一定存在这样的k个相同的数字，那么dp[i][j]就等于 dp[i-pa[i]][j-pb[j]]; 否则，就是找不到这样k个相同的a[i] 和 k个相同的b[i]与其对应。所以，就和a[i]!=b[j]的情况一样了，就等于dp[i][j] = max(dp[i-1][j],dp[i][j-1]);

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<string.h>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
const int N = 1010;
int T,k,n,m;
int dp[N][N];
int a[N],b[N],pa[N],pb[N];
queue<int> q[N];
int main()
{int T; cin>>T;while(T--){cin>>k>>n>>m;memset(dp,0,sizeof(dp));memset(pa,0,sizeof(pa));memset(pb,0,sizeof(pb));rep(i,1,n)cin>>a[i];rep(i,1,m) cin>>b[i];queue<int> q[1005];rep(i,1,n){q[a[i]].push(i);if(q[a[i]].size()==k){pa[i]=q[a[i]].front();q[a[i]].pop();}}rep(i,1,1000) while(!q[i].empty()) q[i].pop();rep(i,1,m){q[b[i]].push(i);if(q[b[i]].size()==k){pb[i]=q[b[i]].front();q[b[i]].pop();}else if(q[b[i]].size()==1) pb[i]=0;}rep(i,1,n)rep(j,1,m){int ans=max(dp[i-1][j],dp[i][j-1]);if(a[i]==b[j] && pa[i]!=0 && pb[j]!=0)ans=max(ans,dp[pa[i]-1][pb[j]-1]+k);dp[i][j]=max(dp[i][j],ans);}cout<<dp[n][m]<<endl;}return 0;
}