DP习题笔记

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;
}