PAT1048. Find Coins（01背包问题动态规划解法）

问题描述：

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 10⁵ coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=10⁵, the total number of coins) and M(<=10³, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V₁ and V₂ (separated by a space) such that V₁ + V₂ = M and V₁ <= V₂. If such a solution is not unique, output the one with the smallest V₁. If there is no solution, output "No Solution" instead.

Sample Input 1:

8 15
1 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 14
1 8 7 2 4 11 15

Sample Output 2:

No Solution

解法描述:

本题可看把需要支付的总额看做背包，每个钱币当做需要放进背包中的物品，钱币的重量和价值均为钱币面额，所以这是一个典型的01背包问题，用动态规划方法依次把各个钱币加入解法中，计算每种总额能拼凑出的最大价值。

代码

 #include<algorithm>
 #include<iostream>
 using namespace std;

 int coins[] = {  };
 int n, m;
 int dp[] = {  };
 int ch[][] = {  };       //ch[n][m]记录每次更新最大价值时，是否选择新加入的钱币

 bool cmp(int a, int b) {
     return a > b;
 }

 int main() {
     cin >> n >> m;
     for (int i = ; i < n; i++)
     {
         cin >> coins[i];
     }
     sort(coins, coins + n, cmp);  //对钱币从大到小排序，因为要输出最小钱币的方案
     for (int i = ; i < n; i++)
     {
         for (int j = m; j >= ; j--)
         {
             if (j >= coins[i]) {
                 if (dp[j] <= dp[j - coins[i]] + coins[i]) {
                     ch[i][j] = ; //若选择了新加入的钱币，则ch[n][m]置为1
                     dp[j] = dp[j - coins[i]] + coins[i];
                 }
             }
         }
     }
     if (dp[m] != m) {            //m总额下的最大价值不是m，则无解决方案
         cout << "No Solution" << endl;
     }
     else {
         for (int i = n - ; i >= ; i--) //从最小的钱币开始倒序遍历
         {
             if (ch[i][m]) {              //若该钱币在拼凑方案中，则总额变为m-coins[i]
                 m -= coins[i];
                 cout << coins[i];
                 if (m == ) {
                     cout << endl;
                 }
                 else {
                     cout << " ";
                 }
             }
         }
     }
     return ;
 }