BZOJ 4198 [Noi2015]荷马史诗

Posted 2020-10-24 ws_zzy

题解：哈夫曼树

先补全节点，以总个数为第一关键字，高度为第二关键字，求哈夫曼树即可

问题：不明白哈夫曼树的原理及证明

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
typedef long long Lint;
const int maxn=100009;

int n,m;
Lint anssum,ansh;

int nn;
Lint sum[maxn<<2];
Lint h[maxn<<2];
struct Point{
	int x;
	bool operator < (const Point &rhs) const{
		if(sum[x]==sum[rhs.x]){
			return h[x]>h[rhs.x];
		}else{
			return sum[x]>sum[rhs.x];
		}
	}
	Point(int xx){
		x=xx;
	}
	Point(){}
};
priority_queue<Point>q;

int main(){
	scanf("%d%d",&n,&m);
	while(!q.empty())q.pop();
	for(int i=1;i<=n;++i){
		scanf("%lld",&sum[i]);
		h[i]=0;
		q.push(Point(i));
	}
	nn=n;
	if(m!=2){
		for(int l=1;;++l){
			if((m-1)*l+1>=n){
				for(int i=n+1;i<=(m-1)*l+1;++i){
					sum[++nn]=0;h[nn]=0;
					q.push(Point(i));
				}
				break;
			}
		}
	}
	
	while(q.size()!=1){
		++nn;
		for(int i=1;i<=m;++i){
			int t=q.top().x;q.pop();
			h[nn]=max(h[nn],h[t]+1);
			sum[nn]+=sum[t];
		}
		ansh=max(ansh,h[nn]);
		anssum=anssum+sum[nn];
		q.push(Point(nn));
	}
	
	printf("%lld\n",anssum);
	printf("%lld\n",ansh);
	return 0;
}