Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3460 Accepted Submission(s): 1092
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
1 5 1 4 2 5 -12 4 -12 1 2 4
Sample Output
2
Source
代码：动态规划求最长增长公共序列 下面展示的是压缩空间的lcs，由于不需要记录顺序，所以这样写，较为简便，如果要记录路径只需要将lcs[]--->换成lcs[][],
然后maxc，变为lcs[][]的上一行即可！
//增长lcs algorithm
#include<stdio.h>
#include<string.h>
#define maxn 505
int aa[maxn],bb[maxn];
int lcs[maxn];
int main()
{
int test,na,nb,i,j,maxc,res;
scanf("%d",&test);
while(test--)
{
scanf("%d",&na);
for(i=;i<=na;i++)
scanf("%d",aa+i);
scanf("%d",&nb);
for(j=;j<=nb;j++)
scanf("%d",bb+j);
memset(lcs,,sizeof(lcs));
for(i=;i<=na;i++)
{
maxc=;
for(j=;j<=nb;j++)
{
if(aa[i]==bb[j]&&lcs[j]<maxc+)
lcs[j]=maxc+;
if(aa[i]>bb[j]&&maxc<lcs[j])
maxc=lcs[j];
}
}
res=;
for(i=;i<=nb;i++)
if(res<lcs[i])res=lcs[i];
printf("%d\n",res);
if(test) putchar();
}
return ;
}