Codeforces 10 D. LCIS

最新推荐文章于 2020-08-09 16:25:15 发布

三级头不怕通

最新推荐文章于 2020-08-09 16:25:15 发布

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本文介绍了解决两个整数序列最长公共上升子序列问题的一种高效算法实现，并提供了详细的AC代码示例，通过动态规划方法求解最长公共上升子序列及其具体数值。

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

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

The sequence s₁, s₂, ..., s_k is called the subsequence of the sequence a₁, a₂, ..., a_n, if there exist such a set of indexes 1 ≤ i₁ < i₂ < ... < i_k ≤ n that a_{i_j} = s_j. In other words, the sequence s can be derived from the sequence a by 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 n space-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.

Examples

input

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

output

3
3 5 6

input

output

2
0 1

我自己一开始写的多用了一维空间

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

ll ans; ll cnt = 0;
const int maxn = 505;
ll a[maxn], b[maxn], dp[maxn][maxn];
ll maxx;
int main(){
    int an, bn;
    cin>>an;
    for (int i=0; i<an; i++){
        cin>>a[i];
    }
    cin>>bn;
    for (int i=0; i<bn; i++){
        cin>>b[i];
    }
    memset(dp, 0, sizeof (dp));
    for (int i=0; i<bn; i++){
        if (a[0] == b[i]){
            dp[0][i] = 1;
        }
    }
    for (int i=1; i<an; i++){
        maxx = 0;
        for (int j=0; j<bn; j++){
            if (a[i] == b[j]) dp[i][j] = maxx+1;
            else dp[i][j] = dp[i-1][j];//没用的步骤
            if (b[j] < a[i]) 
                maxx = max(maxx, dp[i][j]);
        }
    }
    maxx = 0;
    for (int i=0; i<bn; i++){
        maxx = max(maxx, dp[an-1][i]);
    }
    cout<<maxx<<endl;
    return 0;
}

AC代码

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
/*
如何输出路径
dp[i] 为以 bi结尾的最长公共上升字串长度
而且越往后出现的a[i] == b[j], dp[i]肯定越大 所以更新不用判断
所以只要在dp[i]更新时记录是从哪来的 就可以了
*/
ll cnt = 0;
const int maxn = 505;
ll a[maxn], b[maxn], dp[maxn];
int pre[maxn], ans[maxn]; int pos, maxx;
int main(){
    int an, bn;
    cin>>an;
    for (int i=1; i<=an; i++){
        cin>>a[i];
    }
    cin>>bn;
    for (int i=1; i<=bn; i++){
        cin>>b[i];
    }
    memset(dp, 0, sizeof (dp));
    memset(pre, -1, sizeof(pre));
    for (int i=1; i<=an; i++){ 
        pos = 0;
        for (int j=1; j<=bn; j++){///从0开始处理边界麻烦 所以本题从1开始
            if (a[i] == b[j]) dp[j] = dp[pos] +1, pre[j] = pos;
            if (b[j] < a[i] && dp[pos] < dp[j])
                pos = j;
        }
    }
    maxx = -1;
    for (int i=1; i<=bn; i++){
        if (maxx < dp[i]) {
            maxx = dp[i];
            pos = i;
        }
    }

    for (int i=maxx; i; i--){
        ans[i] = b[pos];
        pos = pre[pos];
    }
    cout<<maxx<<endl;
    for (int i=1; i<=maxx; i++){
        printf("%d ", ans[i]);
    }
    puts("");
    return 0;
}