Day44 动态规划part06 完全背包理论基础 518. 零钱兑换 II 377. 组合总和 Ⅳ

动态规划part06 完全背包理论基础 518. 零钱兑换 II 377. 组合总和 Ⅳ

完全背包理论基础

acm可运行代码（先遍历物品再遍历背包，一维dp）

#include
#include
using namespace std;

int Solution(vector<int>& weights,vector<int>& values,int V){
    vector<int> dp(V+1);
    for(int i = 0; i<weights.size();i++){
        for(int j = weights[i]; j<=V;j++){
            dp[j] = max(dp[j],dp[j-weights[i]]+values[i]);
        }
    }
    return dp[V];
}
int main(){
    int N,V;
    cin>>N>>V;
    vector<int> weights(N);
    vector<int> values(N);
    while(N--){
        int weight, value;
        cin>>weight>>value;
        weights.push_back(weight);
        values.push_back(value);
    }
    std::cout << Solution(weights,values,V) << std::endl;
    return 0;
}

先遍历背包再遍历物品，一维dp

#include
#include
using namespace std;

int Solution(vector<int>& weights,vector<int>& values,int V){
    vector<int> dp(V+1);
    for(int j = 0; j <=V;j++){
        for(int i = 0; i<weights.size();i++){
            if(j>=weights[i])dp[j] = max(dp[j],dp[j-weights[i]]+values[i]);
        }
    }
    
    return dp[V];
}
int main(){
    int N,V;
    cin>>N>>V;
    vector<int> weights(N);
    vector<int> values(N);
    while(N--){
        int weight, value;
        cin>>weight>>value;
        weights.push_back(weight);
        values.push_back(value);
    }
    std::cout << Solution(weights,values,V) << std::endl;
    return 0;
}

先遍历背包再遍历物品，二维dp

#include
#include
using namespace std;

int Solution(vector<int>& weights,vector<int>& values,int V){
    int kinds = weights.size();
    vector<vector<int>> dp(kinds,vector<int>(V+1,0));
    for(int i= weights[0];i<=V;i++){
        dp[0][i] = dp[0][i-weights[0]]+values[0];
    }
    for(int i=1;i<weights.size();i++){
        for(int j = 0; j<=V;j++){
            if(j>=weights[i]){
                dp[i][j] = max(dp[i-1][j],dp[i][j-weights[i]]+values[i]);
            }else{
                dp[i][j] = dp[i-1][j];
            }
        }
    }
    return dp[kinds-1][V];
}
int main(){
    int N,V;
    cin>>N>>V;
    vector<int> weights(N);
    vector<int> values(N);
    while(N--){
        int weight, value;
        cin>>weight>>value;
        weights.push_back(weight);
        values.push_back(value);
    }
    std::cout << Solution(weights,values,V) << std::endl;
    return 0;
}

518. 零钱兑换 II

必须先遍历物品，再遍历背包

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

377. 组合总和 Ⅳ

必须先遍历背包，再遍历物品

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target+1);
        dp[0] =1;
        for(int j = 0; j <=target;j++){
            for(int i =0; i<nums.size();i++){
                if(j>=nums[i] && dp[j] < INT_MAX-dp[j-nums[i]]) dp[j] += dp[j-nums[i]];
            }
        }
        return dp[target];
    }
};