1、http://acm.hdu.edu.cn/showproblem.php?pid=1423
参考百度文库http://wenku.baidu.com/view/3e78f223aaea998fcc220ea0.html
2、题目大意：
求两个字符串的最大上升子序列，LCIS解决即可
Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2116 Accepted Submission(s): 638
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
3、代码：
#include<stdio.h>
#include<algorithm>
#include<cstring>
using namespace std;
int main()
{
int t,m,n,a[600],b[600],dp[601][601],i,j,max;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
scanf("%d",&m);
for(i=0;i<m;i++)
scanf("%d",&b[i]);
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++)
{
max=0;
for(j=0;j<m;j++)
{
dp[i+1][j+1]=dp[i][j+1];
if(a[i]>b[j]&&max<dp[i+1][j+1])
max=dp[i+1][j+1];
if(a[i]==b[j])
dp[i+1][j+1]=max+1;
}
}
//printf("%d\n",m);
//for(i=0;i<=m;i++)
//printf("%d\n",dp[n][i]);
max=*max_element(dp[n],dp[n]+m+1);
printf("%d\n",max);
if(t!=0)
printf("\n");
}
return 0;
}