poj 2127 Greatest Common Increasing Subsequence

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 , … , SN of length N is called an increasing subsequence of a sequence A1 , A2 , … , AM of length M if there exist 1 <= i1 < i2 < … < iN <= M such that Sj = Aij for all 1 <= j <= N , and Sj < Sj+1 for all 1 <= j < N .

Input

Each sequence is described with M — its length (1 <= M <= 500) and M integer numbers Ai (-231 <= Ai < 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
5
1 4 2 5 -12
4
-12 1 2 4

Sample Output
2
1 4

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【分析】

题意是给两个数组，求最长公共上升子序列（LCIS）的长度且记录路径

状态：dp[i][j]表示以s1串的前i个字符s2串的前j个字符且以s2[j]为结尾构成的LCIS的长度。

状态转移：当 s1[i-1]!=s2[j-1]时,按照lcs可知由2个状态转移过来,dp[i-1][j],dp[i][j-1],因为dp[i][j]是以s2[j]为结尾构成的LCIS的长度。所以s2[j-1]一定会包含在里面，所以舍去dp[i][j-1],只由dp[i-1][j] 转移过来。当s1[i-1]==s2[j-1]，这时肯定要找前面s1[ii-1]==s2[jj-1]的最长且比s2[j-1]小的状态转移过来.

若s1[i-1]!=s2[j-1] 那么dp[i][j]=dp[i][j-1]

若s1[i-1]==s2[j-1] 那么dp[i][j]=MAX{dp[i-1][k]+1(0 < k < j),s2[k-1] < s2[j-1]};
这里有一步可以优化就是求上一层的最值，可以在更新的同时就求出来，这样就从o(n^3)优化到o(n^2);

mac来记录临时路径

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【代码】

//poj 2127 Greatest Common Increasing Subsequence#include<iostream>#include<cstdio>#include<cstring>#define fo(i,j,k) for(i=j;i<=k;i++)using namespace std;int n,m;int mx,mac,ii,jj,ans,len;int s1[501],s2[501],pre[501][501],dp[501][501],lu[501]; //pre:路径 int main(){    int i,j,k;    scanf("%d",&n);    fo(i,0,n-1)      scanf("%d",&s1[i]);    scanf("%d",&m);    fo(i,0,m-1)      scanf("%d",&s2[i]);    fo(i,1,n)    {        mx=0;        fo(j,1,m)        {            int temp=dp[i][j]=dp[i-1][j];            if(mx<temp && s1[i-1]>s2[j-1])              mx=temp,mac=j;            if(s1[i-1]==s2[j-1])              dp[i][j]=mx+1,pre[i][j]=mac;            if(ans<dp[i][j])              ans=dp[i][j],ii=i,jj=j;        }    }    printf("%d\n",ans);    len=ans;    lu[ans--]=jj-1;    while(ans && ii && jj)    {        if(pre[ii][jj])        {            lu[ans--]=pre[ii][jj]-1;            jj=pre[ii][jj];        }        ii--;    }    fo(i,1,len)      printf("%d ",s2[lu[i]]);    return 0;}