hdu 1423 Greatest Common Increasing Subsequence (最长上升子序列)

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



[align=left]Problem Description[/align]
This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.

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

[align=left]Output[/align]
output print L - the length of the greatest common increasing subsequence of both sequences.

[align=left]Sample Input[/align]

1

5
1 4 2 5 -12
4
-12 1 2 4

[align=left]Sample Output[/align]

2

3、代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[510];
int b[510];
int dp[510][510];
int main()
{
int n1,n2,t,maxx;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n1);
for(int i=1; i<=n1; i++)
{
scanf("%d",&a[i]);
}
scanf("%d",&n2);
for(int i=1; i<=n2; i++)
scanf("%d",&b[i]);
memset(dp,0,sizeof(dp));
for(int i=1; i<=n1; i++)
{
maxx=0;
for(int j=1; j<=n2; j++)
{
dp[i][j]=dp[i-1][j];
if(a[i]>b[j]&&maxx<dp[i-1][j])
maxx=dp[i-1][j];
if(a[i]==b[j])
dp[i][j]=maxx+1;
}
}
maxx=0;
for(int i=1; i<=n2; i++)
if(maxx<dp[n1][i])
maxx=dp[n1][i];
printf("%d\n",maxx);
if(t)
printf("\n");
}
return 0;
}
/*
1

5
1 4 2 5 -12
4
-12 1 2 4
*/