hdoj 1423 Greatest Common Increasing Subsequence(最长上升公共子序列)

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4801    Accepted Submission(s): 1529

Problem Description

This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

 

Input

Each sequence is described with M - its length (1 <= M <= 500) and M integer numbers Ai (-2^31 <= Ai < 2^31) - the sequence itself.

 

Output

output print L - the length of the greatest common increasing subsequence of both sequences.

 

Sample Input

151 4 2 5 -124-12 1 2 4

 

Sample Output

2

 

Source

ACM暑期集训队练习赛（二）

 

#include<cstdio>#include<cmath>#include<cstring>#include<cstdlib>#include<queue>#include<stack>#include<vector>#include<algorithm>using namespace std;#define Inf 0x7fffffffconst int MAX=501;int a[MAX];int b[MAX];int dp[MAX];int main(){int  T;scanf("%d",&T);while(T--){  int n,m,i,j;  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++)     {      int pos=1; for(j=1;j<=m;j++) {  /// 遍历b的同时，找出 j 之前最大的公共子序列所在点 pos，并且要使 b[pos] < a[i]                             /// 这样，当下一句 b[j]==a[i] 时，就可以直接得到 j 点的最大公共子序列了：dp[pos]+1                             if(b[j]<a[i]&&dp[j]+1>dp[pos])    pos=j;   if(b[j]==a[i]&&dp[j]<dp[pos]+1)    dp[j]=dp[pos]+1;           }   }  int ans=-Inf; for(i=1;i<=m;i++)   if(ans<dp[i])    ans=dp[i];printf("%d\n",ans);if(T) printf("\n");} return 0;}