DP习题笔记

最新推荐文章于 2021-07-23 15:55:11 发布

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最新推荐文章于 2021-07-23 15:55:11 发布

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原文链接：http://www.cnblogs.com/lightmonster/p/10657216.html

本文解析了两个经典的动态规划问题：寻找构成整数n所需的最少完全平方数数量及求解给定序列中最长递增子序列的问题。通过具体代码示例展示了如何使用动态规划方法高效解决这两类问题。

Q1：Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, ...) which sum to n.

class Solution {
public:
    int numSquares(int n) {
        if (n == 0) return 0;
        
        vector<int> dp(n+1, 0);
        
        for (int i=0; i<=n; ++i) {
            dp[i] = i;
            for (int j = 2; j<=sqrt(i); ++j) {
                dp[i] = min(dp[i], 1 + dp[i - j*j]);
            }
        }
        
        return dp[n];
    }
};

Q2：A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK<= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

一道求最长上升子序列的DP题，主要思想是找出转移方程dp[i]=max(dp[j]+1) (j<i且a[j]<a[i])。

#include <iostream>
#include <vector>
using namespace std;
int main(int argc, char **argv)
{
    vector<int> a = { 1,2,3,5,9,4,8,10,14,2,5,2 };
    vector<int> dp(a.size(),0);
    dp[0] = 1;
    int ml = 1;
    for (int i = 1; i < a.size(); ++i) {
        for (int j = 0; j < i; ++j) {
            if (a[j] <= a[i])
                dp[i] = max(dp[j] + 1, dp[i]);
        }
        ml = max(ml, dp[i]);
    }
    cout << ml;
    return 0;
}