浅谈(Huffman)树:https://www.cnblogs.com/AKMer/p/10300870.html
题目传送门:https://codeforces.com/problemset/problem/884/D
把分离倒过来就是合并,每次尽量多合并可以保证答案更优,所以问题就转化成了裸的(3)叉(Huffman)树问题。
时间复杂度:(O(nlogn))
空间复杂度:(O(n))
代码如下:
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=2e5+5;
int n;
ll ans;
int read() {
int x=0,f=1;char ch=getchar();
for(;ch<'0'||ch>'9';ch=getchar())if(ch=='-')f=-1;
for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+ch-'0';
return x*f;
}
struct Heap {
int tot;
ll tree[maxn];
void ins(ll v) {
tree[++tot]=v;
int pos=tot;
while(pos>1) {
if(tree[pos]<tree[pos>>1])
swap(tree[pos],tree[pos>>1]),pos>>=1;
else break;
}
}
ll pop() {
ll res=tree[1];
tree[1]=tree[tot--];
int pos=1,son=2;
while(son<=tot) {
if(son<tot&&tree[son|1]<tree[son])son|=1;
if(tree[son]<tree[pos])
swap(tree[son],tree[pos]),pos=son,son=pos<<1;
else break;
}
return res;
}
}T;
int main() {
n=read();
if(n%2==0)T.ins(0);
for(int i=1;i<=n;i++) {
int x=read();
T.ins(x);
}
while(T.tot!=1) {
ll a=T.pop(),b=T.pop(),c=T.pop();
a+=b+c,T.ins(a),ans+=a;
}
printf("%I64d
",ans);
return 0;
}