简单多状态dp【动态规划】

 

目录

 

一、按摩师

二、打家劫舍

 三、删除并获得点数

 四、粉刷房子

 五、买卖股票的最佳时机

 六、买卖股票的最佳时机（含手续费）

七、买卖股票的最佳时机III

 八、买卖股票的最佳时机IV

---

一、按摩师

class Solution {
public:
    int massage(vector<int>& nums) {
 
        int n =nums.size();
        if(n == 0) return 0;
        vector<int> f(n);
        auto g = f;
        f[0] = nums[0];
 
        for(int i = 1;i < n;i++){
            f[i] = g[i-1] + nums[i];
            g[i] = max(f[i-1],g[i-1]);
        }
        return max(f[n-1],g[n-1]);
    }
};

二、打家劫舍

class Solution {
public:
    int rob1(vector<int>& nums,int l,int r) {
        if(l>r) return 0;
        int n =nums.size();
        if(n == 0) return 0;
        vector<int> f(n);
        auto g = f;
        f[l] = nums[l];
 
        for(int i = l;i <= r;i++){
            f[i] = g[i-1] + nums[i];
            g[i] = max(f[i-1],g[i-1]);
        }
        return max(f[r],g[r]);
    }
 
    int rob(vector<int>& nums) {
        int n = nums.size();
 
        int ret1 = rob1(nums,2,n-2)+nums[0];
        int ret2 = rob1(nums,1,n-1);
        return max(ret1,ret2);
    }
};

 三、删除并获得点数

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        int n = nums.size();
        const int N = 10001;
        int arr[N] = {0}; 
        for(auto e : nums)
        {
            arr[e] += e;
        }
 
        vector<int> f(N);
        auto g = f;
        for(int i = 1;i < N;i++){
            f[i] = g[i-1] + arr[i];
            g[i] = max(f[i-1],g[i-1]);
        }
        return max(f[N-1],g[N-1]);
    }
};

 四、粉刷房子

 

class Solution {
public:
    int minCost(vectorint>>& costs) {
        int n = costs.size();
        vectorint>> dp(n+1,vector<int>(3));
        for(int i = 1;i <= n;i++)
        {
            dp[i][0] = costs[i-1][0] + min(dp[i-1][1],dp[i-1][2]);
            dp[i][1] = costs[i-1][1] + min(dp[i-1][0],dp[i-1][2]);
            dp[i][2] = costs[i-1][2] + min(dp[i-1][0],dp[i-1][1]);
        }
        return min(dp[n][0],min(dp[n][1],dp[n][2]));
    }
};

 五、买卖股票的最佳时机

 

 

  

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vectorint>> dp(n,vector<int>(3));
        dp[0][0] = -prices[0];
        
        for(int i = 1;i < n;i++){
            dp[i][0] = max(dp[i-1][0],dp[i-1][1] - prices[i]);
            dp[i][1] = max(dp[i-1][1],dp[i-1][2]);
            dp[i][2] = dp[i-1][0]+prices[i];
        }
 
        return max(dp[n-1][1],dp[n-1][2]);
    }
};

 六、买卖股票的最佳时机（含手续费）

上一题用的是二维数组的第二维来表示多种状态，是因为状态比较多，如果像此题只有两种状态，就可以用两个函数，本质上是一样的。 

 

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int n = prices.size();
        vector<int> f(n);
        auto g = f;
        f[0] = -prices[0];
        
        for(int i = 1;i < n;i++){
            f[i] = max(f[i-1],g[i-1] - prices[i]);
            g[i] = max(g[i-1],f[i-1]+prices[i]- fee);
        }
        return g[n-1];
    }
};

七、买卖股票的最佳时机III

 

 

class Solution {
public:
    const int INF = 0x3f3f3f3f;
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vectorint>> f(n,vector<int>(3,-INF));
        auto g = f;
        f[0][0] = -prices[0];
        g[0][0] = 0;
        for(int i = 1;i < n;i++){
            for(int j = 0;j < 3;j++){
                f[i][j] = max(f[i-1][j],g[i-1][j] - prices[i]);
                g[i][j] = g[i-1][j];
                if(j >= 1)
                    g[i][j] = max(g[i-1][j],f[i-1][j-1]+prices[i]);
            }
        }   
        int ret = 0;
        for(int i = 0;i < 3;i++){
            ret = max(ret,g[n-1][i]);
        }
        return ret;
    }
};

 八、买卖股票的最佳时机IV

 

class Solution {
public:
    int maxProfit(int k, vector<int>& prices) {
        int  n = prices.size();
        k = min(k,n/2); 
        const int INF = 0x3f3f3f3f;
        vectorint>> f(n,vector<int>(k+1,-INF));//注意是k+1
        auto g = f;
        f[0][0] = -prices[0];
        g[0][0] = 0;
        for(int i = 1;i < n;i++){
            for(int j = 0;j <= k;j++){
                f[i][j] = max(f[i-1][j],g[i-1][j]-prices[i]);
                g[i][j] = g[i-1][j];
                if(j >= 1)
                    g[i][j] = max(g[i-1][j],f[i-1][j-1]+prices[i]);
            }
        }
        int ret = 0;
        for(int j = 0;j <= k;j++){
            ret = max(ret,g[n-1][j]);
        }
        return ret;
    }
};