动态规划-背包问题

0-1背包

N件物品，背包最大容量为V, 第i件物品的费用为w[i],价值为v[i]
使用f[i][j]表示在容量为j，在前i件物品中(包括i)选择物品所得到的最大价值
递推方程为f[i][j] = max(f[i-1][j], f[i-1][j - w[i]] + v[i]) 在是否选择第i件物品取最大值
从后往前更新就可使用一维数组简化f[j] = max(f[j], f[j-w[i]] + v[i])
416. Partition Equal Subset Sum

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int sum = accumulate(nums.begin(), nums.end(), 0);
        return sum & 1 ? false : subSum(nums, sum >> 1);
    }
    
    bool subSum(vector<int>& nums, int s){
        bool dp[s + 1] = {false};
        dp[0] = true;
        for(int n : nums){
            for(int i = s; i >=n; i--){
                dp[i] = dp [i] || dp[i - n];
            }
        }
        
        return dp[s];
    }
    
};

彻底背包

每种物品无限件, 递推方程为
f[i][v]=max(f[i-1][v-k*c[i]]+k*w[i]|0<=k*c[i]<=v)
322. Coin Change

//超时
class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
   
        if(amount == 0) return 0;
        vector<vector<int>> f(n+1, vector<int>(amount+1, amount + 1));
        
        for(int i = 0; i < n; i++){
            f[i][0] = 0;
            for(int j = 1; j <= amount; j++){
                for(int k = 0; k * coins[i] <= j; k++){
                    f[i+1][j] = min(f[i+1][j], f[i][j - k * coins[i]] + k);
                }
            }
        }
        
        return f[n][amount] < amount + 1 ? f[n][amount] : -1;
    }
};

优化时间，三重循环变为两重循环, 注意这两重循环可交换

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
   
        if(amount == 0) return 0;
        vector<vector<int>> f(n+1, vector<int>(amount+1, amount + 1));
        for(int i = 0; i <= n; i++) f[i][0] = 0;
        for(int j = 1; j <= amount; j++){
            for(int i = 0; i < n; i++){
                
                if(j - coins[i] >= 0)
                    f[i+1][j] = min(f[i][j], f[i+1][j - coins[i]] + 1);
                else f[i+1][j] = f[i][j];
                
            }
        }
        
        return f[n][amount] < amount + 1 ? f[n][amount] : -1;
    }
};

优化空间，二维数组变为一维数组

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
   
        if(amount == 0) return 0;
        vector<int> f(amount+1, amount + 1);
        f[0] = 0;
        for(int j = 1; j <= amount; j++){
            for(int i = 0; i < n; i++){
          
                if(j - coins[i] >= 0)
                    f[j] = min(f[j], f[j - coins[i]] + 1);

            }
        }
        
        return f[amount] < amount + 1 ? f[amount] : -1;
    }
};

518. Coin Change 2
作题的时候仍是要写个二维的验证一下

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        
        for(int i = 0; i < n; i++){
            for(int j = coins[i]; j <= amount; j++){
                dp[j] += dp[j - coins[i]];
                // dp[i+1][j] = dp[i][j] + dp[i+1][j - coins[i]]
            }
        }
        
        return dp[amount];
    }
};

多重背包

初始化问题

理解合法状态，要看清题目中说的是正好放满背包，仍是最多放满背包
前者对应dp[i][0] = 0, dp[i][j] = INF(j != 0, 不是合法状态)，后者对应dp[i][0] = 0(全是合法状态)

参考

背包九讲