hdu 1423 Greatest Common Increasing Subsequence(最长公共递增子序列LICS)

Greatest Common Increasing Subsequence

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

参考：/article/10367218.html

用一个二维数组dp[i][j]表示a序列的前i项，b序列的前j项，并且以b[j]结束的LCIS的长度。

则有：

当a[i] != b[j]时，dp[i][j]=dp[i-1][j];

当a[i] == b[j]时，dp[i][j]=max(dp[i-1][k])+1,1<=k<=j-1 && b[j]>b[k]; //求dp[i-1]行中的最长，并且保证新增的数大于已经有的数

但是这样的效率为O(n^3)。 不难发现，求dp[i-1][k]的时候可以通过一个变量temp来解决，当k从1遍历到j的同时，可以不断更新temp的值，如果a[i]>b[j]时，令temp=max(temp,dp[i-1][j])；如果a[i]==b[j]时，令dp[i][j]=max+1。

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

int n,m,a[505],b[505],dp[505][505];
int LICS()
{

int Max;

for (int i=1;i<=n; i++)
{
Max=0;
for (int j=1;j<=m; j++)
{
dp[i][j]=dp[i-1][j];
if (a[i]>b[j] && Max<dp[i-1][j])
Max=dp[i-1][j];
if (a[i]==b[j])
dp[i][j]=Max+1;
}
}
Max=0;
for (int i=1; i<=m; i++)
Max=max(Max,dp
[i]);

return Max;
}

int main()
{
int i,t;
scanf("%d",&t);
while(t--)
{
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]);
printf("%d\n",LICS());
if(t)
printf("\n");
}

return 0;
}

上面的虽然可以解决，但是二维浪费空间较大，我们注意到在LICS函数中有一句dp[i][j] = dp[i-1][j]，这证明dp数组前后没有变化！于是可以优化成一维数组！

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

int n,m,a[505],b[505],dp[505];
int LCIS()
{

int Max;
memset(dp, 0, sizeof(dp));
for (int i=1;i<=n; i++)
{
Max=0;
for (int j=1;j<=m; j++)
{
if (a[i]>b[j] && Max<dp[j])
Max=dp[j];
if (a[i]==b[j])
dp[j]=Max+1;
}
}
Max=0;
for (int i=1; i<=m; i++)
Max=max(Max,dp[i]);

return Max;
}

int main()
{
int i,t;
scanf("%d",&t);
while(t--)
{
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]);
printf("%d\n",LCIS());
if(t)
printf("\n");
}

return 0;
}