CF10D LCIS

题意翻译

求两个串的最长公共上升子序列。

题目描述

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 s s can be derived from the sequence a a 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.

输入输出格式

输入格式：

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

输出格式：

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.

输入输出样例

输入样例#1： 

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

输出样例#1： 

3
3 5 6 

输入样例#2： 

5
1 2 0 2 1
3
1 0 1

输出样例#2： 

2
0 1 

Solution：

　　本题线性DP。

　　直接把两个常见的dp揉合，定义状态$f[i][j]$表示匹配到$A$串第$i$位并以$B$串第$j$位结尾的LCIS长度。

　　那么不难得到状态转移方程：$f[i][j]=max(f[i-1][k]+1),a[i]==b[j]&&a[i]>b[k]$。

　　显然可以滚掉第一维，然后若直接转移复杂度是$n^3$的，我们可以优化掉枚举决策$k$的循环，在状态转移完后存下最大的满足条件的$f[k]$作为下次转移的决策，显然满足最优性，这样的时间复杂度是$O(n^2)$的。

　　输出方案就只需要在转移时顺带记录下$b$串每个位置的$pre$就好了。

代码：

/*Code by 520 -- 10.20*/
#include<bits/stdc++.h>
#define il inline
#define ll long long
#define RE register
#define For(i,a,b) for(RE int (i)=(a);(i)<=(b);(i)++)
#define Bor(i,a,b) for(RE int (i)=(b);(i)>=(a);(i)--)
using namespace std;
const int N=505;
int n,a[N],m,b[N],f[N],pre[N];
int stk[N],top,ans;

int main(){
    scanf("%d",&n); For(i,1,n) scanf("%d",a+i);
    scanf("%d",&m); For(i,1,m) scanf("%d",b+i);
    For(i,1,n) {
        int pos=0;
        For(j,1,m){
            if(a[i]==b[j]) f[j]=f[pos]+1,pre[j]=pos;
            if(a[i]>b[j]&&f[pos]<f[j]) pos=j;    
        }
    }
    int pos=0;
    For(i,1,m) if(f[i]>f[pos]) pos=i;
    printf("%d
",f[pos]);
    while(f[pos]--) stk[++top]=b[pos],pos=pre[pos];
    while(top) printf("%d ",stk[top--]);
    return 0;
}