动态规划
Dynamic Programming
简写为 DP,是运筹学的一个分支,是求解决策过程最优化的过程。20世纪50年代初,美国数学家贝尔曼(R.Bellman)等人在研究多阶段决策过程的优化问题时,提出了著名的最优化原理,从而创立了动态规划。动态规划的应用极其广泛,包括工程技术、经济、工业生产、军事以及自动化控制等领域,并在背包问题、生产经营问题、资金管理问题、资源分配问题、最短路径问题和复杂系统可靠性问题等中取得了显著的效果。
动态规划算法的基本步骤包括:
- 确定状态:确定需要求解的状态,并将其表示为变量。
- 确定状态转移方程:根据问题的特定约束条件和目标函数,确定状态之间的转移关系,并将其表示为数学公式。
- 初始化:为初始状态赋初值,并将其表示为初始条件。
- 递推计算:根据状态转移方程,使用循环依次计算各个状态的解,并将其保存在数组或表中。
- 求解最终结果:根据问题的目标,从计算得到的解中得出最终结果。
动态规划算法可以用于解决各种问题,例如最短路径问题、背包问题、最长公共子序列问题等。在实现动态规划算法时,需要根据具体问题的特点进行设计和调整,以确保算法的正确性和效率。
适用条件
任何思想方法都有一定的局限性,超出了特定条件,它就失去了作用。同样,动态规划也并不是万能的。适用动态规划的问题必须满足最优化原理和无后效性。
最优化原理(最优子结构性质)
最优化原理可这样阐述:一个最优化策略具有这样的性质,不论过去状态和决策如何,对前面的决策所形成的状态而言,余下的诸决策必须构成最优策略。简而言之,一个最优化策略的子策略总是最优的。一个问题满足最优化原理又称其具有最优子结构性质 [8] 。
无后效性
将各阶段按照一定的次序排列好之后,对于某个给定的阶段状态,它以前各阶段的状态无法直接影响它未来的决策,而只能通过当前的这个状态。换句话说,每个状态都是过去历史的一个完整总结。这就是无后向性,又称为无后效性 [8] 。
子问题的重叠性
动态规划算法的关键在于解决冗余,这是动态规划算法的根本目的。动态规划实质上是一种以空间换时间的技术,它在实现的过程中,不得不存储产生过程中的各种状态,所以它的空间复杂度要大于其他的算法。选择动态规划算法是因为动态规划算法在空间上可以承受,而搜索算法在时间上却无法承受,所以我们舍空间而取时间。
真题举例(1)
44. 通配符匹配
给定一个字符串 (s
) 和一个字符模式 (p
) ,实现一个支持 '?'
和 '*'
的通配符匹配。
'?' 可以匹配任何单个字符。'*' 可以匹配任意字符串(包括空字符串)。
两个字符串完全匹配才算匹配成功。
说明:
-
s
可能为空,且只包含从a-z
的小写字母。 -
p
可能为空,且只包含从a-z
的小写字母,以及字符?
和*
。
示例 1:
输入:s = "aa" p = "a" 输出: false 解释: "a" 无法匹配 "aa" 整个字符串。
示例 2:
输入:s = "aa" p = "*" 输出: true 解释: '*' 可以匹配任意字符串。
示例 3:
输入:s = "cb" p = "?a" 输出: false 解释: '?' 可以匹配 'c', 但第二个 'a' 无法匹配 'b'。
示例 4:
输入:s = "adceb" p = "*a*b" 输出: true 解释: 第一个 '*' 可以匹配空字符串, 第二个 '*' 可以匹配字符串 "dce".
示例 5:
输入:s = "acdcb" p = "a*c?b" 输出: false
代码1:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
bool isMatch(string s, string p)
{
vector<vector<bool>> dp(s.size() + 1, vector<bool>(p.size() + 1));
dp[0][0] = 1;
for (int j = 1; j <= p.size(); j++)
{
dp[0][j] = dp[0][j - 1] && p[j - 1] == '*';
}
for (int i = 1; i <= s.size(); i++)
{
for (int j = 1; j <= p.size(); j++)
{
if (p[j - 1] == '*')
{
dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
}
else
{
dp[i][j] = (s[i - 1] == p[j - 1] || p[j - 1] == '?') && dp[i - 1][j - 1];
}
}
}
return dp[s.size()][p.size()];
}
};
int main()
{
Solution s;
cout << s.isMatch("aa", "a") << endl;
cout << s.isMatch("aa", "*") << endl;
cout << s.isMatch("cb", "?a") << endl;
cout << s.isMatch("adceb", "*a*b") << endl;
cout << s.isMatch("acdcb", "a*c?b") << endl;
return 0;
}
代码2:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
bool isMatch(string s, string p)
{
int m = s.size();
int n = p.size();
vector<vector<int>> dp(m + 1, vector<int>(n + 1));
dp[0][0] = true;
for (int i = 1; i <= n; i++)
{
if (p[i - 1] == '*')
dp[0][i] = true;
else
break;
}
for (int i = 1; i <= m; i++)
{
for (int j = 1; j <= n; j++)
{
if (p[j - 1] == '*')
{
dp[i][j] |= dp[i][j - 1];
dp[i][j] |= dp[i - 1][j];
}
else
{
if (p[j - 1] == '?' || s[i - 1] == p[j - 1])
{
dp[i][j] |= dp[i - 1][j - 1];
}
}
}
}
return dp[m][n];
}
};
int main()
{
Solution s;
cout << s.isMatch("aa", "a") << endl;
cout << s.isMatch("aa", "*") << endl;
cout << s.isMatch("cb", "?a") << endl;
cout << s.isMatch("adceb", "*a*b") << endl;
cout << s.isMatch("acdcb", "a*c?b") << endl;
return 0;
}
代码3:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
bool isMatch(string s, string p)
{
if (p.empty())
return s.empty();
if (s.empty())
{
if (p[0] == '*')
return isMatch(s, p.substr(1));
else
return false;
}
if (p[0] == '*')
return isMatch(s, p.substr(1)) || isMatch(s.substr(1), p);
else
return (s[0] == p[0] || p[0] == '?') && isMatch(s.substr(1), p.substr(1));
}
};
int main()
{
Solution s;
cout << s.isMatch("aa", "a") << endl;
cout << s.isMatch("aa", "*") << endl;
cout << s.isMatch("cb", "?a") << endl;
cout << s.isMatch("adceb", "*a*b") << endl;
cout << s.isMatch("acdcb", "a*c?b") << endl;
return 0;
}
代码4:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool isMatch(string s, string p) {
if (s == "" && p == "" || (s == "" && p == "*"))
return true;
if (s == p)
return true;
int lens = s.length();
int lenp = p.length();
bool questionm = false, starm = false;
for (int k = 0; k < lenp; k++) {
if (p[k] == '?')
questionm = true;
if (p[k] == '*')
starm = true;
}
if (lenp != lens && questionm == false && starm == false)
return false;
int i = 0, j = 0;
int mstar = 0, sstar = -1;
while (i < lens) {
if (j < lenp && p[j] == '*') {
mstar = i;
sstar = j;
j += 1;
} else if (j < lenp && (s[i] == p[j] || p[j] == '?')) {
i++;
j++;
} else if (sstar != -1) {
mstar += 1;
j = sstar + 1;
i = mstar;
} else
return false;
}
while (j < lenp) {
if (p[j] != '*')
return false;
j++;
}
return true;
}
};
int main()
{
Solution s;
cout << s.isMatch("aa", "a") << endl;
cout << s.isMatch("aa", "*") << endl;
cout << s.isMatch("cb", "?a") << endl;
cout << s.isMatch("adceb", "*a*b") << endl;
cout << s.isMatch("acdcb", "a*c?b") << endl;
return 0;
}
62. 不同路径
一个机器人位于一个 m x n
网格的左上角 (起始点在下图中标记为 “Start” )。
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为 “Finish” )。问总共有多少条不同的路径?
示例 1:
输入:m = 3, n = 7 输出:28
示例 2:
输入:m = 3, n = 2 输出:3 解释:从左上角开始,总共有 3 条路径可以到达右下角。 1. 向右 -> 向下 -> 向下 2. 向下 -> 向下 -> 向右 3. 向下 -> 向右 -> 向下
示例 3:
输入:m = 7, n = 3 输出:28
示例 4:
输入:m = 3, n = 3 输出:6
提示:
1 <= m, n <= 100
- 题目数据保证答案小于等于
2 * 10^9
代码1:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int uniquePaths(int m, int n)
{
int N = m + n - 2;
int M = m < n ? m - 1 : n - 1;
long ans = 1;
for (int i = 1; i <= M; i++)
ans = ans * (N - i + 1) / i;
return ans;
}
};
int main()
{
Solution s;
cout << s.uniquePaths(3, 7) << endl;
cout << s.uniquePaths(3, 2) << endl;
cout << s.uniquePaths(7, 3) << endl;
cout << s.uniquePaths(3, 3) << endl;
return 0;
}
代码2:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int uniquePaths(int m, int n)
{
if (m <= 0 || n <= 0)
{
return 0;
}
vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
for (int i = 0; i < m; i++)
{
dp[i][0] = 1;
}
for (int i = 0; i < n; i++)
{
dp[0][i] = 1;
}
for (int i = 1; i < m; i++)
{
for (int j = 1; j < n; j++)
{
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
};
int main()
{
Solution s;
cout << s.uniquePaths(3, 7) << endl;
cout << s.uniquePaths(3, 2) << endl;
cout << s.uniquePaths(7, 3) << endl;
cout << s.uniquePaths(3, 3) << endl;
return 0;
}
代码3:
#include <bits/stdc++.h>
using namespace std;
typedef vector<int> BigInt;
class Solution
{
public:
int uniquePaths(int m, int n)
{
if (m == 0 || n == 0)
return 0;
if (m == 1 || n == 1)
return 1;
int m_ = m - 1 + n - 1;
int n_ = n - 1;
BigInt a = fac(m_);
int result = 0;
for (int i = n_; i >= 1; i--)
a = div(a, i);
for (int i = m_ - n_; i >= 1; i--)
a = div(a, i);
int k = a.size() - 1;
while (a[k] == 0)
k--;
for (int i = k; i >= 0; i--)
result = result * 10 + a[i];
return result;
}
BigInt fac(int n)
{
BigInt result;
result.push_back(1);
for (int factor = 1; factor <= n; ++factor)
{
long long carry = 0;
for (auto &item : result)
{
long long product = item * factor + carry;
item = product % 10;
carry = product / 10;
}
if (carry > 0)
{
while (carry > 0)
{
result.push_back(carry % 10);
carry /= 10;
}
}
}
return result;
}
BigInt div(BigInt a, int d)
{
int b = 0;
BigInt result;
int len = a.size();
for (int i = len - 1; i >= 0; i--)
{
b = b * 10 + a[i];
result.insert(result.begin(), b / d);
b = b % d;
}
return result;
}
};
int main()
{
Solution s;
cout << s.uniquePaths(3, 7) << endl;
cout << s.uniquePaths(3, 2) << endl;
cout << s.uniquePaths(7, 3) << endl;
cout << s.uniquePaths(3, 3) << endl;
return 0;
}
代码4:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int uniquePaths(int m, int n) {
vector<vector<int>> path(m, vector<int>(n, 0));
for (int i = 0; i < n; i++)
path[0][i] = 1;
for (int i = 0; i < m; i++)
path[i][0] = 1;
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
path[i][j] = path[i - 1][j] + path[i][j - 1];
return path[m - 1][n - 1];
}
};
int main()
{
Solution s;
cout << s.uniquePaths(3, 7) << endl;
cout << s.uniquePaths(3, 2) << endl;
cout << s.uniquePaths(7, 3) << endl;
cout << s.uniquePaths(3, 3) << endl;
return 0;
}
63. 不同路径 II
一个机器人位于一个 m x n 网格的左上角 (起始点在下图中标记为“Start” )。
机器人每次只能向下或者向右移动一步。机器人试图达到网格的右下角(在下图中标记为“Finish”)。
现在考虑网格中有障碍物。那么从左上角到右下角将会有多少条不同的路径?
网格中的障碍物和空位置分别用 1
和 0
来表示。
示例 1:
输入:obstacleGrid = [[0,0,0],[0,1,0],[0,0,0]] 输出:2 解释:3x3 网格的正中间有一个障碍物。从左上角到右下角一共有 2 条不同的路径: 1. 向右 -> 向右 -> 向下 -> 向下 2. 向下 -> 向下 -> 向右 -> 向右
示例 2:
输入:obstacleGrid = [[0,1],[0,0]] 输出:1
提示:
m == obstacleGrid.length
n == obstacleGrid[i].length
1 <= m, n <= 100
-
obstacleGrid[i][j]
为0
或1
代码1:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid)
{
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
int p[m][n];
int k = 0;
while (k < m && obstacleGrid[k][0] != 1)
p[k++][0] = 1;
while (k < m)
p[k++][0] = 0;
k = 0;
while (k < n && obstacleGrid[0][k] != 1)
p[0][k++] = 1;
while (k < n)
p[0][k++] = 0;
for (int i = 1; i < m; i++)
for (int j = 1; j < n; j++)
{
if (obstacleGrid[i][j] == 1)
p[i][j] = 0;
else
p[i][j] = p[i - 1][j] + p[i][j - 1];
}
return p[m - 1][n - 1];
}
};
int main()
{
Solution s;
vector<vector<int>> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
obstacleGrid = {{0,1},{0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
return 0;
}
代码2:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid)
{
int m = obstacleGrid.size(), n = obstacleGrid[0].size();
vector<vector<int>> dp(m, vector<int>(n, 0));
for (int i = 0; i < m && obstacleGrid[i][0] != 1; i++)
{
dp[i][0] = 1;
}
for (int i = 0; i < n && obstacleGrid[0][i] != 1; i++)
{
dp[0][i] = 1;
}
for (int i = 1; i < m; i++)
{
for (int j = 1; j < n; j++)
{
if (obstacleGrid[i][j] != 1)
{
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
};
int main()
{
Solution s;
vector<vector<int>> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
obstacleGrid = {{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
return 0;
}
代码3:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid)
{
if (obstacleGrid.size() == 0 || obstacleGrid[0].size() == 0)
return 0;
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
vector<vector<int>> info(m, vector<int>(n, 0));
for (int i = 0; i < m; ++i)
{
if (obstacleGrid[i][0] == 1)
{
for (int j = i; j < m; j++)
{
info[j][0] = 0;
}
break;
}
else
info[i][0] = 1;
}
for (int i = 0; i < n; ++i)
{
if (obstacleGrid[0][i] == 1)
{
for (int j = i; j < n; ++j)
{
info[0][j] = 0;
}
break;
}
else
info[0][i] = 1;
}
for (int i = 1; i < m; ++i)
{
for (int j = 1; j < n; ++j)
{
if (obstacleGrid[i][j] == 1)
{
info[i][j] = 0;
}
else
{
info[i][j] = info[i - 1][j] + info[i][j - 1];
}
}
}
return info[m - 1][n - 1];
}
};
int main()
{
Solution s;
vector<vector<int>> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
obstacleGrid = {{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
return 0;
}
代码4:
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int uniquePathsWithObstacles(vector<vector<int>> &obstacleGrid) {
int m = obstacleGrid.size(), n = obstacleGrid[0].size();
if (obstacleGrid[0][0] == 1 || obstacleGrid[m - 1][n - 1] == 1)
return 0;
vector<vector<int>> dp(m, vector<int>(n, 0));
dp[0][0] = 1;
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 0)
dp[i][0] = dp[i - 1][0];
}
for (int i = 1; i < n; i++) {
if (obstacleGrid[0][i] == 0)
dp[0][i] = dp[0][i - 1];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] == 0)
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
};
int main()
{
Solution s;
vector<vector<int>> obstacleGrid = {{0,0,0},{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
obstacleGrid = {{0,1,0},{0,0,0}};
cout << s.uniquePathsWithObstacles(obstacleGrid) << endl;
return 0;
}
64. 最小路径和
给定一个包含非负整数的 m x n
网格 grid
,请找出一条从左上角到右下角的路径,使得路径上的数字总和为最小。
说明:每次只能向下或者向右移动一步。
示例 1:
输入:grid = [[1,3,1],[1,5,1],[4,2,1]] 输出:7 解释:因为路径 1→3→1→1→1 的总和最小。
示例 2:
输入:grid = [[1,2,3],[4,5,6]] 输出:12
提示:
m == grid.length
n == grid[i].length
1 <= m, n <= 200
0 <= grid[i][j] <= 100
代码1:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int minPathSum(vector<vector<int>> &grid)
{
if (grid.size() == 0)
return 0;
int m = grid.size();
int n = grid[0].size();
vector<vector<int>> m_memo = vector<vector<int>>(m + 1, vector<int>(n + 1, 0));
for (int i = n - 1; i >= 0; --i)
m_memo[m - 1][i] = grid[m - 1][i] + m_memo[m - 1][i + 1];
for (int j = m - 1; j >= 0; --j)
m_memo[j][n - 1] = grid[j][n - 1] + m_memo[j + 1][n - 1];
for (int i = m - 2; i >= 0; --i)
{
for (int j = n - 2; j >= 0; --j)
{
m_memo[i][j] = grid[i][j] + min(m_memo[i][j + 1], m_memo[i + 1][j]);
}
}
return m_memo[0][0];
}
};
int main()
{
Solution s;
vector<vector<int>> grid = {{1,3,1},{1,5,1},{4,2,1}};
cout << s.minPathSum(grid) << endl;
grid = {{1,2,3},{4,5,6}};
cout << s.minPathSum(grid) << endl;
return 0;
}
代码2:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int minPathSum(vector<vector<int>> &grid)
{
int row = grid.size();
int column = grid[0].size();
for (int i = 1; i < column; ++i)
{
grid[0][i] = grid[0][i - 1] + grid[0][i];
}
for (int i = 1; i < row; ++i)
{
grid[i][0] = grid[i - 1][0] + grid[i][0];
}
for (int i = 1; i < row; ++i)
{
for (int j = 1; j < column; ++j)
{
int temp = grid[i - 1][j] > grid[i][j - 1] ? grid[i][j - 1] : grid[i - 1][j];
grid[i][j] = grid[i][j] + temp;
}
}
return grid[row - 1][column - 1];
}
};
int main()
{
Solution s;
vector<vector<int>> grid = {{1,3,1},{1,5,1},{4,2,1}};
cout << s.minPathSum(grid) << endl;
grid = {{1,2,3},{4,5,6}};
cout << s.minPathSum(grid) << endl;
return 0;
}
代码3:
#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
int minPathSum(vector<vector<int>> &grid)
{
int row = grid.size();
int col = grid[0].size();
vector<int> f(col, 0);
for (int i = 0; i < row; ++i)
{
f[0] = f[0] + grid[i][0];
for (int j = 1; j < col; ++j)
{
if (i == 0)
f[j] = f[j - 1] + grid[i][j];
else
f[j] = min(f[j - 1], f[j]) + grid[i][j];
}
}
return f[col - 1];
}
};
int main()
{
Solution s;
vector<vector<int>> grid = {{1,3,1},{1,5,1},{4,2,1}};
cout << s.minPathSum(grid) << endl;
grid = {{1,2,3},{4,5,6}};
cout << s.minPathSum(grid) << endl;
return 0;
}
代码4: 文章来源:https://www.toymoban.com/news/detail-625304.html
#include <bits/stdc++.h>
using namespace std;
class Solution {
private:
int m, n;
int memo[100][100];
public:
int minPathSum(vector<vector<int>> &grid) {
m = grid.size(), n = grid[0].size();
for (int i = 0; i < m; i++) {
memset(memo[i], -1, sizeof(int) * n);
}
return dfs(grid, 0, 0);
}
int dfs(vector<vector<int>> &grid, int r, int c) {
if (r < 0 || r >= m || c < 0 || c >= n)
return 1000000;
if (memo[r][c] != -1)
return memo[r][c];
if (r == m - 1 && c == n - 1) {
memo[r][c] = grid[m - 1][n - 1];
return memo[r][c];
}
int right = dfs(grid, r, c + 1);
int down = dfs(grid, r + 1, c);
memo[r][c] = min(right, down) + grid[r][c];
return memo[r][c];
}
};
int main()
{
Solution s;
vector<vector<int>> grid = {{1,3,1},{1,5,1},{4,2,1}};
cout << s.minPathSum(grid) << endl;
grid = {{1,2,3},{4,5,6}};
cout << s.minPathSum(grid) << endl;
return 0;
}
续:https://hannyang.blog.csdn.net/article/details/132091605文章来源地址https://www.toymoban.com/news/detail-625304.html
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