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  • 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
    ",LICS());
            if(t)
            printf("
    ");
        }
    
        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
    ",LICS());
            if(t)
                printf("
    ");
        }
    
        return 0;
    }
    


     

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  • 原文地址:https://www.cnblogs.com/jiangu66/p/3225983.html
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