Codeforces Beta Round #10 D. LCIS（DP&LCIS）

D. LCIS

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

This problem differs from one which was on the online contest.

The sequence a1, a2, ..., an is
called increasing, if ai < ai + 1 for i < n.

The sequence s1, s2, ..., sk is
called the subsequence of the sequence a1, a2, ..., an,
if there exist such a set of indexes 1 ≤ i1 < i2 < ... < ik ≤ n that aij = sj.
In other words, the sequence s can be derived from the sequence aby
crossing out some elements.

You are given two sequences of integer numbers. You are to find their longest common increasing subsequence, i.e. an increasing sequence of maximum length that is the subsequence of both sequences.

Input

The first line contains an integer n (1 ≤ n ≤ 500)
— the length of the first sequence. The second line contains nspace-separated integers from the range [0, 109] —
elements of the first sequence. The third line contains an integer m (1 ≤ m ≤ 500)
— the length of the second sequence. The fourth line contains m space-separated integers from the range [0, 109] —
elements of the second sequence.

Output

In the first line output k — the length of the longest common increasing subsequence. In the second line output the subsequence itself. Separate
the elements with a space. If there are several solutions, output any.

Sample test(s)

input

7
2 3 1 6 5 4 6
4
1 3 5 6

output

3
3 5 6

input

5
1 2 0 2 1
3
1 0 1

output

2
0 1

题意：

给你长度分别为n,m(1<=n,m<=500)的序列。要你求这两个序列的最长公共上升子序列。

思路:

最长公共子序列做过。最长上升子序列也做过。可是这题时最长公共上升子序列。。。解法肯定还是dp拉。

感觉这题的dp方程的思想非常不错，体现了一种加强约束的思想。dp[i][j]表示。处理完A序列的前i个，且上升序列以B序列的B[j]结尾的最长子序列。感觉这个把状态体现以什么结尾是非常不错的思想。然后转移显而易见了。

if(A[i]==B[j])

dp[i][j]=dp[i-1][k];//k是小于j且B[k]<B[j]

else

dp[i][j]=dp[i-1][j];

我们能够i,j循环这样就能够省掉找k的时间。复杂度O(n*m)

具体见代码：

#include<bits/stdc++.h>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
typedef long long ll;
int dp[550][550],A[550],B[550],path[550][550],n,m;
bool print(int x)
{
    if(!x)
        return false;
    if(print(path[n][x]))
        printf(" %d",B[x]);
    else
        printf("%d",B[x]);
    return true;
}
int main()
{
    int i,j,tp,ans,pos;
    while(~scanf("%d",&n))
    {
        for(i=1;i<=n;i++)
            scanf("%d",&A[i]);
        scanf("%d",&m);
        for(i=1;i<=m;i++)
            scanf("%d",&B[i]);
        memset(dp,0,sizeof dp);
        for(i=1;i<=n;i++)
        {
            tp=pos=0;
            for(j=1;j<=m;j++)
            {
                dp[i][j]=dp[i-1][j];
                path[i][j]=path[i-1][j];
                if(A[i]==B[j]&&tp+1>dp[i][j])
                    dp[i][j]=tp+1,path[i][j]=pos;
                if(B[j]<A[i]&&dp[i-1][j]>tp)
                    tp=dp[i-1][j],pos=j;
            }
        }
        ans=1;
        for(i=1;i<=m;i++)
            if(dp[n][i]>dp[n][ans])
                ans=i;
        printf("%d\n",dp[n][ans]);
        if(dp[n][ans])
            print(ans);
        printf("\n");
    }
    return 0;
}

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