HDOJ 题目1423 Greatest Common Increasing Subsequence（LICS）

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4693    Accepted Submission(s): 1496

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）

 

Recommend

lcy   |   We have carefully selected several similar problems for you:  1422 1400 1424 1401 1404 

 

ac代码

#include<stdio.h>#include<string.h>#define max(a,b) (a>b?a:b)int a[550],b[550],n,m,dp[550];int lics(){    int ans=0,i,j;    for(i=1;i<=n;i++)    {        int ma=0;        for(j=1;j<=m;j++)        {            int temp=ma;            if(b[j]<a[i]&&dp[j]>ma)                ma=dp[j];            if(b[j]==a[i])                dp[j]=temp+1;            ans=max(ans,dp[j]);        }    }    return ans;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        //int n,m;        int i,j;        scanf("%d",&n);        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        scanf("%d",&m);        for(j=1;j<=m;j++)            scanf("%d",&b[j]);        printf("%d\n",lics());        if(t)            printf("\n");    }}