HDU1423：Greatest Common Increasing Subsequence(LICS)

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

题意：求最长递增公共子序列的长度

思路：直接模板

#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,i,j;
    memset(dp,0,sizeof(dp));
    for(i = 1; i<=n; i++)
    {
        MAX = 0;
        for(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(i = 1; i<=m; i++)
        if(MAX<dp[n][i])
            MAX = dp[n][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 <string.h>
#include <algorithm>
using namespace std;

int a[505],b[505],dp[505],n,m;

int LICS()
{
    int i,j,MAX;
    memset(dp,0,sizeof(dp));
    for(i = 1; i<=n; i++)
    {
        MAX = 0;
        for(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(i = 1; i<=m; i++)
        if(MAX<dp[i])
            MAX = dp[i];
    return MAX;
}

int main()
{
    int t,i;
    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;
}