这道题可以用二叉堆做。 每次从二叉堆里拿最小的两堆果子,合并后,记录成本,并且把新的堆加入。循环直至仅剩1堆即可。
#include<bits/stdc++.h>
using namespace std;
using LL=long long;
#define SMA
#ifdef SMA
#define sym <
#else
#define sym >
#endif
template<typename datatype,const int size>class bheap{//sma root example
public:
int n;
int heap[size];
bheap(){
n=0;
}
void up(int p){
while(p>1){
if(heap[p] sym heap[p/2]){//small root failed ,son < par
swap(heap[p],heap[p/2]);
p/=2;
}else{
break;
}
}
}
void insert(int val){
heap[++n]=val;
up(n);
}
void down(int p){
int son=p*2;//point to left child
while(son<=n){//in legal area
if(son<n&&heap[son]>heap[son+1]){//choose smaller son
++son;
}//choose lchild or right child
if(heap[son]<heap[p]){//parent > smallest son
//!fail to adapt to small root!
swap(heap[son],heap[p]);
p=son;son=2*p;
//update
}else{
break;
}
}
}
void extract(){
heap[1]=heap[n--];
down(1);
}
int getTop(){
return heap[1];
}
void remove(int k){
heap[k]=heap[n--];
up(k),down(k);
}
};
int main(){
#ifdef LOCAL
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
int n;
cin>>n;
bheap<int,10005> sheap;
int tem;
for(int in=0;in<n;++in){
cin>>tem;
sheap.insert(tem);
}
long long ans=0;
while(sheap.n>=3){
int temval=0;
if(sheap.heap[2]<sheap.heap[3]){
temval=sheap.heap[2]+sheap.heap[1];
ans+=temval;
sheap.remove(2);
sheap.extract();
sheap.insert(temval);
}else{
temval=sheap.heap[3]+sheap.heap[1];
ans+=temval;
sheap.remove(3);
sheap.extract();
sheap.insert(temval);
}
}
if(sheap.n==2){
int temval=sheap.heap[2]+sheap.heap[1];
ans+=temval;
sheap.remove(2);
sheap.extract();
sheap.insert(temval);
}
cout<<ans;
}