hdu 1423 Greatest Common Increasing Subsequence（不连续子序列）

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4841    Accepted Submission(s): 1543

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

151 4 2 5 -124-12 1 2 4

 

Sample Output

2
最长上升公共子序列，dp，，好好学学啊，唉2015,4,25，，下边是大神的解释  * DP（最长公共递增非连续子序列Greatest Common Increasing Subsequence）  *   设数组f[i][j]代表第一个序列前i个数与第二个序列前j个数的 最长公共递增非连续子序列（GCIS） 的长度。  *   而数组tmp[i][j]则代表第一个序列前i个数与第二个序列前j个数的GCIS的最大数（即最后一个数）  *                        max{f[i-1][j-1],f[i-1][j],f[i][j-1]}    (当a[i]!=a[j])  *                       /  *   状态转移方程：f[i][j]=  *                       \  *                        max{f[i-1][j-1]+1(满足a[i]>tmp[i-1][j-1]),         f[i-1][j-1],f[i-1][j]+1(满足a[i]>tmp[i-1][j]),         f[i-1][j],f[i][j-1]+1(满足a[i]>tmp[i][j-1]),f[i][j-1]}   (当a[i]==a[j])  */  
#include<stdio.h>#include<string.h>#define M 550#define max(a,b) (a>b? a:b)int dp[M][M],a[M],b[M];int main(){int t,m1,m2,i,j,k;scanf("%d",&t);while(t--){scanf("%d",&m1);for(i=1;i<=m1;i++)scanf("%d",&a[i]);scanf("%d",&m2);for(i=1;i<=m2;i++)scanf("%d",&b[i]);memset(dp,0,sizeof(dp));for(i=1;i<=m1;i++){k=0;for(j=1;j<=m2;j++){dp[i][j]=dp[i-1][j];if(b[j]<a[i]&&k<dp[i-1][j])k=dp[i-1][j];if(b[j]==a[i])dp[i][j]=k+1;}}k=-1;for(i=1;i<=m2;i++)if(k<dp[m1][i])k=dp[m1][i];printf("%d\n",k);if(t!=0)printf("\n");}return 0;}