2019 ICPC Asia Taipei-Hsinchu Regional Problem K Length of Bundle Rope (贪心,优先队列)

• 题意:有\(n\)堆物品,每次可以将两堆捆成一堆,新堆长度等于两个之和,每次消耗两个堆长度之和的长度,求最小消耗使所有物品捆成一堆.

• 题解:贪心的话,每次选两个长度最小的来捆,这样的消耗一定是最小的,但是我们需要一个容器来存这些数,这时候很明显要用到优先队列(小根堆),我们将所有元素入队,每次取前两个捆,捆完后入队即可.

• 代码:

  #include <iostream>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
   
  int t;
  int n;
   
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);
  	cin>>t;
  	 while(t--){
  	 	cin>>n;
  	 	priority_queue <int,vector<int>,greater<int>> q;
  	 	for(int i=1;i<=n;++i){
  	 		int x;
  	 		cin>>x;
  	 		q.push(x);
  	 	}
  	 	int res=0;
  	 	while(q.size()>=2){
  	 		int tmp1=q.top();
  	 		q.pop();
  	 		int tmp2=q.top();
  	 		q.pop();
  	 		res+=tmp1+tmp2;
  	 		q.push(tmp1+tmp2);
  	 	}
  	 	cout<<res<<endl;
  	 }
      return 0;
  }