动态规划思想
通过状态转移方程将大问题分解为小问题，找到小问题的最优解，然后基于此小问题的最优状态获得另一问题的最优解。具体做法是将各个小问题的最优解存入数组中，然后基于其来获取当前问题的最优解。
在应用中就是找规律，将此规律用状态转移方程来表示。
以下通过例题来学习此算法！！！
题目一：
有1,3,5三种面值的硬币，找到总值为16元的最少硬币数量？
获得d(0)、d(1)、d(2)小问题的最优解
d(0) = 0
d(2) = d(2-1)+1 = d(1)+1 = 1+1 = 2
d(3) = d(3-3)+1 = d(0)+1 = 1
d(3) = d(3-1)+1 = d(2)+1 = 3
d(4) = d(4-3)+1 = d(1)+1 = 2
状态转移方程：
d(x) = min( d(x-v)+1 )
#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { //面值 int v[3] = { 1,3,5 }; //索引为总值，元素为硬币数量 vector<int> arr1; arr1.push_back(0); for (int i = 1; i <= 16; i++) { //所有面值对应的硬币数量 vector<int> arr2; for (int j = 0; j < 3; j++) { //总值需大于等于面值 if (i - v[j] >= 0) { arr2.push_back(arr1[i - v[j]] + 1); } } //获得当前总值对应的最少硬币数量 sort(arr2.begin(), arr2.end()); arr1.push_back(arr2[0]); } for (int i = 0; i < arr1.size(); i++) cout << arr1[i] << " "; cout<<"最终结果为："<< arr1[arr1.size()-1]<<endl; return 0; } 题目二：
给定长度为N的序列，获得最长的非降序列?
4,3,5,7,5,8
d(0) = 1
d(1) = 1
d(2) = d(1)+1 = 2
d(3) = d(2)+1 = 3
d(4) = 1
d(5) = d(4)+1 = 2
状态转移方程：
if arr[i]>arr[i-1]
else
d(i) = 1
#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { int seq[6] = { 4,3,5,7,5,8 }; vector<int > arr; arr.push_back(1); for (int i = 1; i < 6; i++) { if (seq[i] > seq[i - 1]) arr.push_back(arr[i - 1] + 1); else arr.push_back(1); } for (int i = 0; i < arr.size(); i++) cout << arr[i] << " "; cout << endl; sort(arr.begin(), arr.end()); cout <<"最优结果为："<< arr[arr.size() - 1] << endl; return 0; } 题目三：
上台阶有两种方式，一种是一次上一阶，另一种是一次上两阶，若有N个台阶，则有多少种走法?
获得d(n-2)、d(n-1)小问题的最优解，然后再此基础上再走一步或两步台阶
d(1) = d(0) = 1
d(2) = d(2-2) + d(2-1) = 1+1 = 2
d(3) = d(3-2) + d(3-1) = 1+2 = 3
d(4) = d(4-2) + d(4-1) = 2+4 = 5
状态转移方程
d(n) = d(n-2) + d(n-1)
#include <iostream> #include <vector> #include <algorithm> using namespace std; int main() { int n = 4; vector<int> arr; arr.push_back(1); arr.push_back(1); if (n == 1) cout << "走法：" << n << endl; for (int i = 2; i < n; i++) arr.push_back(arr[i - 2] + arr[i - 1]); for (int i = 0; i < arr.size(); i++) cout << arr[i] << " "; cout << endl; cout << "走法：" << arr[arr.size() - 1] << endl; return 0; } 题目四：
给定二维数组，从左上角开始到右下角仅能向右或向下移动，获得最优路径？
2,6,4,6
1,5,7,0
3,7,4,7
1,3,0,2
创建二维数组，获得每个位置在当前位置处的最优路径
d(0,0) = a(0,0) = 2
d(0,1) = a(0,1) + d(0,0) = 8
...
d(2,0) = a(2,0) + d(1,0) = 3+3 = 6
...
d(1,2) = a(1,2) + min{d(0,2), d(1,1)} = 7+8 = 15
...
状态转移方程
d(i,j) = a(i,j) + min{d(i-1,j) + d(i,j-1)}
#include <iostream> using namespace std; int min(int a, int b) { return a < b ? a : b; } int main() { int arr[4][4] = { 2,6,4,6,1,5,7,0,3,7,4,7,1,3,0,2 }; int d[4][4] = {}; //将当前位置的最优解存储在数组中 for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { if (i == 0 && j == 0) d[i][j] = arr[i][j]; else if (i == 0 && j != 0) d[i][j] = arr[i][j] + d[i][j - 1]; else if (i != 0 && j == 0) d[i][j] = arr[i][j] + d[i - 1][j]; else d[i][j] = arr[i][j] + min(d[i][j-1],d[i-1][j]); } } for (int i = 0; i < 4; i++) { for (int j = 0; j < 4; j++) { cout << d[i][j] << " "; } cout << endl; } cout << "最终解：" << d[3][3] << endl; return 0; }