poj2127Greatest Common Increasing Subsequence【LICS】

Description

You are given two sequences of integer numbers. Write a program to determine their common increasing subsequence of maximal possible length.
Sequence S₁ , S₂ , . . . , S_N of length N is called an increasing subsequence of a sequence A₁ , A₂ , . . . , A_M of length M if there exist 1 <= i₁ < i₂ < . . . < i_N <= M such that S_j = A_ij for all 1 <= j <= N , and S_j < S_j+1 for all 1 <= j < N .

Input

Each sequence is described with M --- its length (1 <= M <= 500) and M integer numbers A_i (-2³¹ <= A_i < 2³¹ ) --- the sequence itself.

Output

On the first line of the output file print L --- the length of the greatest common increasing subsequence of both sequences. On the second line print the subsequence itself. If there are several possible answers, output any of them.

Sample Input

51 4 2 5 -124-12 1 2 4

Sample Output

21 4

Source

Northeastern Europe 2003, Northern Subregion

裸的LICS加上输出路径就把自己搞懵了；_； 一心想着记录父节点可以递归输出，然而远不用那么麻烦，最长递增公共子序列嘛==而且是二维的（废话），多加一个二维数组记录哈希了的前一个对应坐标，根据这个坐标记录每步的走法，这么说来，我原来递归的写法就只差一个数组呗，等回头补上

/*********poj21272016.2.121168K141MSC++1188B*********/#include <iostream>#include<cstdio>#include<cstring>using namespace std;int n,m,x[505],y[505],dp[505],pos[505][505],ans[505];void solve(){    memset(dp,0,sizeof(dp));    memset(pos,0,sizeof(pos));    for(int i=1;i<=n;i++)    {        int tmp=0;        memcpy(pos[i],pos[i-1],sizeof(pos[0]));///        for(int j=1;j<=m;j++)        {            if(x[i]>y[j]&&dp[j]>dp[tmp]) tmp=j;            if(x[i]==y[j]&&dp[tmp]+1>dp[j])            {                dp[j]=dp[tmp]+1;                pos[i][j]=i*(m+1)+tmp;            }        }    }    int maxn=0;    for(int i=1;i<=m;i++) if(dp[maxn]<dp[i]) maxn=i;    int i=m*n+n+maxn;    for(int j=dp[maxn];j>0;j--)    {        ans[j]=y[i%(m+1)];        i=pos[i/(m+1)][i%(m+1)];    }    printf("%d\n",dp[maxn]);    for(i=1;i<=dp[maxn];i++)    {        if(i!=1) printf(" %d",ans[i]);        else printf("%d",ans[i]);    } puts("");}int main(){   // freopen("cin.txt","r",stdin);    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++) scanf("%d",&x[i]);        scanf("%d",&m);        for(int i=1;i<=m;i++) scanf("%d",&y[i]);        solve();    }    return 0;}