AcWing 刷题计划：0x10 基本数据结构——AcWing 148. 合并果子

1.题目描述：

2.题意：
略。

3.思路：
贪心。（哈夫曼树） 每次选择最小的两堆进行合并。由于每次都是两个堆的数量的加和。所以我们如果把两个大的数再合并，则体力会越来越大。

4.代码：

//AcWing 148. 合并果子
//#include<bits/stdc++.h>
//#pragma GCC optimize(3,"Ofast","inline")
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
//#include<random>
#include<cstdlib>
#include<ctime>
#include<map>
#include<set>
#include<stack>
#include<queue>
#define FAST ios::sync_with_stdio(false)
#define DEV_RND ((int)rand()*RAND_MAX+rand())
#define RND(L,R) (DEV_RND%((R)-(L)+1)+(L))
#define abs(a) ((a)>=0?(a):-(a))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,n,a) for(int i=n-1;i>=a;--i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define li inline
#define re register
using namespace std;
//typedef uniform_int_distribution<int> RNDI;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef double db;
typedef long long ll;
typedef long double ld;
const int maxn = 1e5+5;
const int maxm = 100000+5;
const int mod = 1e9+7;
const int inf=0x3f3f3f3f;
const double eps = 1e-7;
const double pi=acos(-1);
//int dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
//li int f(int x){return x==par[x]?par[x]:par[x]=f(par[x]);}
//mt19937 eng(time(0));
//li int RND(int L,int R){RNDI rnd(L,R);return rnd(eng);}
li ll lowbit(ll x){return x&(-x);}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);
li ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res%MOD;}
li ll qmul(ll a,ll b,ll MOD=mod){return (a*b-(ll)((long double)a/MOD*b)*MOD+MOD)%MOD;}
li ll inv(ll x,ll p){return qpow(x,p-2,p);}
li ll jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}
db f(db x){return x;}
li db sim(db l,db r){return (f(l)+4.*f((l+r)/2.)+f(r))*(r-l)/6.;}
db asr(db l,db r,db ans,db eps){db m=l+(r-l)/2.,L=sim(l,m),R=sim(m,r);return fabs(L+R-ans)<=15.*eps?L+R+(L+R-ans)/15.:asr(l,m,L,eps/2)+asr(m,r,R,eps/2);}
db asr(db l,db r,db eps){return asr(l,r,sim(l,r),eps);}
namespace IO
{li ll read(){ll x=0,sign=1;char c=getchar();while(c>'9'||c<'0') {if(c=='-') sign=-1;c=getchar();}while('0'<=c&&c<='9') x=x*10+c-'0',c=getchar();return x*sign;}template<typename T>li void write(T x,char t='\n'){if(x<0){x=-x;putchar('-');};static int sta[25];int top=0;do{sta[top++]=x%10;}while(x/=10);while(top) putchar(sta[--top]+'0');putchar(t);}
}
using namespace IO;
/*-------------head-------------*/
//
int n,m;
priority_queue<int,VI,greater<int>> q;
li void solve()
{rep(i,0,n) q.push(read());int cnt=0;while(sz(q)!=1){int x1=q.top();q.pop();int x2=q.top();q.pop();cnt+=x1+x2;q.push(x1+x2);}write(cnt);//puts("");
}
int main()
{//srand(time(0));//freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);//for(int QwQ=read();QwQ;QwQ--) solve();while(~scanf("%d",&n)) solve();return 0;
}