PAT 1068 Find More Coins (30)

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 must pay the exact amount. Since she has as many as 10^4^ 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 some 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 (<=10^4^, the total number of coins) and M(<=10^2^, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the face values V~1~ <= V~2~ <= ... <= V~k~ such that V~1~ + V~2~ + ... + V~k~ = M. All the numbers must be separated by a space, and there must be no extra space at the end of the line. If such a solution is not unique, output the smallest sequence. If there is no solution, output "No Solution" instead.

Note: sequence {A[1], A[2], ...} is said to be "smaller" than sequence {B[1], B[2], ...} if there exists k >= 1 such that A[i]=B[i] for all i < k, and A[k] < B[k].

Sample Input 1:

8 9
5 9 8 7 2 3 4 1

Sample Output 1:

1 3 5

Sample Input 2:

4 8
7 2 4 3

Sample Output 2:

No Solution

 

题目大意：给出一个数组，找出和等于目标数的子集， 如果子集不唯一， 则输出最小的子集；

思路：先对数组进行排序，在进行dfs()遍历， 因为数组以及实现排序， 如果找到和为目标数的子集， 则一定是最小的子集。把结果保存在vecotr<int> ans中， 如果ans为空，则表示没有满足要求的子集

注意点：对临时数组temp压栈， 应该在循环中，而不是在进入dfs()的地方， 否则第二个测试点不能通过；

 

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> temp, ans, v;
bool flag=false;
int k;
void dfs(int node, int val){
  if(flag) return;
  if(val==0){
    ans = temp;
    flag= true;
    return;
  }else if(val<0){
    temp.pop_back();
    return;
  }
  for(int i=node; i<v.size()&&!flag; i++){
    temp.push_back(v[i]);
    dfs(i+1, val-v[i]);
  }
  temp.pop_back();
}
int main(){
  int n, i;
  scanf("%d %d", &n, &k);
  v.resize(n);
  for(i=0; i<n; i++) scanf("%d", &v[i]);
  sort(v.begin(), v.end());
  dfs(0, k);
  if(!flag) printf("No Solution
");
  for(i=0; i<ans.size(); i++){
    if(i==0) printf("%d", ans[i]);
    else printf(" %d", ans[i]);
  }
  
  return 0;
}