代码随想录 - 背包问题

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    vector<int> weight = {1, 3, 4};
    vector<int> value = {15, 20, 30};
    int bagWeight = 4;
    // 01背包 如果先遍历背包 后遍历物品  那就是每个包只会装一个物品
    for(int j = bagWeight; j >= 0; j--) {
        for(int i = 0; i < weight.size(); i++) { // 遍历物品
         // 遍历背包容量
         if (j>=weight[i])
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i]);
        }
        for (auto i : dp) cout << i << " ";
        cout << endl;
    }

上面的dp
0 0 0 0 30
0 0 0 20 30
0 0 15 20 30
0 15 15 20 30
0 15 15 20 30

完全背包的排列和组合的区别
https://programmercarl.com/0518.零钱兑换II.html#思路

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class Solution {
public:
    vector<int> coins= {1, 2, 5};
    int amount = 5;

    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int i = 0; i < coins.size(); i++) { // 遍历物品
            for (int j = coins[i]; j <= amount; j++) { // 遍历背包
                dp[j] += dp[j - coins[i]];
            }
        }
        return dp[amount];
    }
};

上面是组合， dp是
j = 1, coin[i] = 1, dp: 1 1 0 0 0 0
j = 2, coin[i] = 1, dp: 1 1 1 0 0 0
j = 3, coin[i] = 1, dp: 1 1 1 1 0 0
j = 4, coin[i] = 1, dp: 1 1 1 1 1 0
j = 5, coin[i] = 1, dp: 1 1 1 1 1 1
j = 2, coin[i] = 2, dp: 1 1 2 1 1 1
j = 3, coin[i] = 2, dp: 1 1 2 2 1 1
j = 4, coin[i] = 2, dp: 1 1 2 2 3 1
j = 5, coin[i] = 2, dp: 1 1 2 2 3 3
j = 5, coin[i] = 5, dp: 1 1 2 2 3 4

Copy
class Solution {
public:
    vector<int> coins= {1, 2, 5};
    int amount = 5;

    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int j = 0; j <= amount; j++) { // 遍历背包容量
            for (int i = 0; i < coins.size(); i++) { // 遍历物品
                if (j - coins[i] >= 0) {
                    dp[j] += dp[j - coins[i]];
                    for (auto i : dp) cout << i << " ";
                    cout << endl;
                }
            }
        }
        return dp[amount];
    }
};

上面是排列，dp是
j = 1, coin[i] = 1, dp: 1 1 0 0 0 0
j = 2, coin[i] = 1, dp: 1 1 1 0 0 0
j = 2, coin[i] = 2, dp: 1 1 2 0 0 0
j = 3, coin[i] = 1, dp: 1 1 2 2 0 0
j = 3, coin[i] = 2, dp: 1 1 2 3 0 0
j = 4, coin[i] = 1, dp: 1 1 2 3 3 0
j = 4, coin[i] = 2, dp: 1 1 2 3 5 0
j = 5, coin[i] = 1, dp: 1 1 2 3 5 5
j = 5, coin[i] = 2, dp: 1 1 2 3 5 8
j = 5, coin[i] = 5, dp: 1 1 2 3 5 9