hdu 1423 Greatest Common Increasing Subsequence 最大公共上升子序列 DP

Greatest Common Increasing Subsequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4590    Accepted Submission(s): 1462

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

 

最近一直在做DP题。。现在的进步是，可以看得懂别人的题解了，。

上一个百度文库的文章讲的还是非常不错的。最长公共上升子序列(LCIS)的平方算法

下面是用O(n^3)的算法。还会更新O(n^2)的算法 。

#include <stdio.h>#include <string.h>#define MAX 600int dp[MAX][MAX] ;int max(int a ,int b){return a>b?a:b ;}int main(){int num1[MAX],num2[MAX], t;scanf("%d",&t);while(t--){int n , m ;scanf("%d",&n);for(int i = 1 ; i <= n ; ++i)scanf("%d",&num1[i]);scanf("%d",&m);for(int i = 1 ; i <= m ; ++i)scanf("%d",&num2[i]);memset(dp,0,sizeof(dp)) ;int ans  = 0 ;for(int i = 1 ; i <= n ; ++i){for(int j = 1 ; j <= m ; ++j){if(num1[i]!=num2[j])dp[i][j] = dp[i-1][j];if(num1[i] == num2[j]){dp[i][j] = 1 ;for(int k = 1 ; k < j ; ++k){if(num2[j]>num2[k])dp[i][j] = max(dp[i][j],dp[i-1][k]+1) ;}}ans = max(ans,dp[i][j]) ;}}printf("%d\n",ans) ;if(t != 0){printf("\n") ;}}return 0 ;}

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晚上更新：

捣鼓了O(n^2)+一维数组的算法，，不得不佩服搞出这个算法的人，神牛啊！！我看都这么费劲，人家却能发明出来。哎，深刻的感受到了这个世界恶意

o(╯□╰)o

看代码：

#include <stdio.h>#include <string.h>#define MAX 505 int getMax(int a , int b){return a>b?a:b ;}int main(){int num1[MAX],num2[MAX],dp[MAX],t;scanf("%d",&t);while(t--){int n , m ;scanf("%d",&n);for(int i = 1 ; i <= n ; ++i)scanf("%d",&num1[i]);scanf("%d",&m);for(int i = 1 ; i <= m ; ++i)scanf("%d",&num2[i]);memset(dp,0,sizeof(dp)) ;int ans  = 0 ;for(int i = 1 ; i <= n ; ++i){int max = 0 ;for(int j = 1 ; j <= m ; ++j){if(num1[i]>num2[j] && max < dp[j])max = dp[j] ;if(num1[i] == num2[j]){dp[j] = max+1 ;}ans = getMax(ans,dp[j]) ;}}printf("%d\n",ans) ;if(t!=0){printf("\n") ;}}return 0 ;}