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  • 最长公共上升子序列

    Greatest Common Increasing Subsequence
    
    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3362    Accepted Submission(s): 1060
    
    
    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

    自己写的公式,min也是自己想到的,哈哈,太高兴了

    #define _CRT_SECURE_NO_DEPRECATE
    #include<iostream>
    #include<string>
    #include<algorithm>
    #define MAX 502
    using namespace std;
    int dp[MAX][MAX];
    int a[MAX], b[MAX];
    int main()
    {
        int T, m, n, min;
        cin >> T;
        while (T--){
            cin >> m;
            for (int i = 0; i < m; i++){
                cin >> a[i];
            }
            cin >> n;
            for (int i = 0; i < n; i++){
                cin >> b[i];
            }
            min = INT_MIN;
            fill(dp[0], dp[0] + n, 0);
            for (int i = 0; i < m; i++){
                for (int j = 0; j < n; j++){
                    if (a[i] == b[j] && a[i]>min){
                        min = a[i];
                        dp[i + 1][j + 1] = dp[i][j] + 1;
                    }
                    else{
                        dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);
                    }
                }
            }
            cout << dp[m][n] << endl;
            if (T){
                cout << endl;
            }
        }
        
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/littlehoom/p/3551319.html
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