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
题解:输入两个数字串a,b,求出它们最长公共递增子序列的长度。
代码:
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[550][550];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m,a[550],b[550],i,j;
memset(dp,0,sizeof(dp));
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]);
for(i=1; i<=n; i++)
{
int ans=0;
for(j=1; j<=m; j++)
{
dp[i][j]=dp[i-1][j];
if(a[i]>b[j]&&ans<dp[i-1][j])
ans=dp[i-1][j];
if(a[i]==b[j])
dp[i][j]=ans+1;
}
}
int sum=0;
for(i=1; i<=m; i++)
if(sum<dp[n][i])
sum=dp[n][i];
printf("%d\n",sum);
if(t)
printf("\n");
}
return 0;
}