HDU 1423 Greatest Common Increasing Subsequence
Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8060 Accepted Submission(s): 2600
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
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <vector> #include <iomanip> #include <cmath> #include <ctime> #include <map> #include <set> using namespace std; #define lowbit(x) (x&(-x)) #define max(x,y) (x>y?x:y) #define min(x,y) (x<y?x:y) #define MAX 100000000000000000 #define MOD 1000000007 #define pi acos(-1.0) #define ei exp(1) #define PI 3.141592653589793238462 #define INF 0x3f3f3f3f3f #define mem(a) (memset(a,0,sizeof(a))) typedef long long ll; int dp[1006]; int a[1006]; int b[1006]; int n,m,t; int main() { scanf("%d",&t); while(t--) { scanf("%d",&n); int maxn; for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } scanf("%d",&m); for(int i=1;i<=m;i++) { scanf("%d",&b[i]); } memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { maxn=0; for(int j=1;j<=m;j++) { if(a[i]>b[j]) maxn=max(dp[j],maxn); if(a[i]==b[j]) dp[j]=maxn+1; } } maxn=0; for(int i=1;i<=m;i++) maxn=max(dp[i],maxn); printf("%d\n",maxn); if(t) printf("\n"); } return 0; }