硬币更改DP算法打印所有组合

时间:2015-10-26 18:40:18

标签: c++ algorithm vector dynamic-programming coin-change

这里很好地描述了经典的硬币更改问题:http://www.algorithmist.com/index.php/Coin_Change

在这里,我不仅要知道有多少种组合,还要打印出所有这些组合。我在我的实现中在该链接中使用相同的DP算法,但不是在DP[i][j] = count的DP表中记录多少组合,而是将组合存储在表中。所以我正在为这个DP表使用3D矢量。

我试图改进我的实现,注意到在查找表时,只需要最后一行的信息,所以我真的不需要总是存储整个表。

然而,我改进的DP解决方案似乎仍然很慢,所以我想知道我的实现中是否存在一些问题,或者可以进行更多优化。谢谢!

您可以直接运行代码:

#include <iostream>
#include <stdlib.h>
#include <iomanip>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;

int main(int argc, const char * argv[]) {       

    int total = 10; //total amount
    //available coin values, always include 0 coin value
    vector<int> values = {0, 5, 2, 1}; 
    sort(values.begin(), values.end()); //I want smaller coins used first in the result 

    vector<vector<vector<int>>> empty(total+1); //just for clearing purpose
    vector<vector<vector<int>>> lastRow(total+1);
    vector<vector<vector<int>>> curRow(total+1);


    for(int i=0; i<values.size(); i++) {


        for(int curSum=0; curSum<=total; curSum++){
            if(curSum==0) {
                //there's one combination using no coins               
                curRow[curSum].push_back(vector<int> {}); 

            }else if(i==0) {
                //zero combination because can't use coin with value zero

            }else if(values[i]>curSum){
                //can't use current coin cause it's too big, 
                //so total combination for current sum is the same without using it
                curRow[curSum] = lastRow[curSum];

            }else{
                //not using current coin
                curRow[curSum] = lastRow[curSum];
                vector<vector<int>> useCurCoin = curRow[curSum-values[i]];

                //using current coin
                for(int k=0; k<useCurCoin.size(); k++){

                    useCurCoin[k].push_back(values[i]);
                    curRow[curSum].push_back(useCurCoin[k]);
                }               
            }    
        }        

        lastRow = curRow;
        curRow = empty;
    } 

    cout<<"Total number of combinations: "<<lastRow.back().size()<<endl;
    for (int i=0; i<lastRow.back().size(); i++) {
        for (int j=0; j<lastRow.back()[i].size(); j++) {
            if(j!=0)
                cout<<" ";
            cout<<lastRow.back()[i][j];
        }
        cout<<endl;
    }
    return 0;
}

1 个答案:

答案 0 :(得分:1)

您似乎复制了太多的向量:至少最后else可以重写为

// not using current coin
curRow[curSum] = lastRow[curSum];
const vector<vector<int>>& useCurCoin = curRow[curSum - values[i]]; // one less copy here

// using current coin
for(int k = 0; k != useCurCoin.size(); k++){
    curRow[curSum].push_back(useCurCoin[k]);
    curRow[curSum].back().push_back(values[i]); // one less copy here too.
}

即使清除curRow = empty;是可读的,也可能会产生分配。 最好创建一个功能

void Clean(vector<vector<vector<int>>>& vecs)
{
    for (auto& v : vecs) {
        v.clear();
    }
}