Greatest Common Increasing Subsequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3649 Accepted Submission(s): 1147
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
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #define clr(x,y) memset(x,y,sizeof(x)) 5 using namespace std; 6 int a[600],b[600],ans[600],an,bn; 7 int work() 8 { 9 memset(ans,0,sizeof(ans)); 10 int i,j,t,maxa,maxan=0; 11 for(i=1; i<=an; i++) 12 { 13 maxa=0; 14 for(j=1; j<=bn; j++) 15 { 16 if(a[i]>b[j]&&ans[j]>maxa)maxa=ans[j]; 17 if(a[i]==b[j]&&ans[j]<maxa+1) 18 ans[j]=maxa+1; 19 } 20 } 21 for(i=1;i<=bn;i++) 22 if(ans[i]>maxan)maxan=ans[i]; 23 return maxan; 24 } 25 int main() 26 { 27 int t,i; 28 scanf("%d",&t); 29 while(t--) 30 { 31 scanf("%d",&an); 32 for(i=1; i<=an; i++)scanf("%d",&a[i]); 33 scanf("%d",&bn); 34 for(i=1; i<=bn; i++)scanf("%d",&b[i]); 35 printf("%d\n",work()); 36 if(t)printf("\n"); 37 } 38 }
Source
