poj_3253

源地址：http://poj.org/problem?id=3253

霍夫曼树的应用。每次选择两个长度最短的拿来切，最后的结果一定是最小的。

我们可以用优先队列来解决，这个模板可以快速解决这种问题。不过注意要用long long。

#include<stdio.h>
#include<iostream>
#include<string>
#include<string.h>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<time.h>
#include<queue>
#include<stack>
#include<iterator>
#include<math.h>
#include<stdlib.h>
#include<limits.h>
#include<map>
//#define ONLINE_JUDGE
#define eps 1e-8
#define INF 0x7fffffff
#define FOR(i,a) for((i)=0;i<(a);(i)++)
#define MEM(a) (memset((a),0,sizeof(a)))
#define sfs(a) scanf("%s",a)
#define sf(a) scanf("%d",&a)
#define sfI(a) scanf("%I64d",&a)
#define pf(a) printf("%d\n",a)
#define pfI(a) printf("%I64d\n",a)
#define pfs(a) printf("%s\n",a)
#define sfd(a,b) scanf("%d%d",&a,&b)
#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)
#define for1(i,a,b) for(int i=(a);i<b;i++)
#define for2(i,a,b) for(int i=(a);i<=b;i++)
#define for3(i,a,b)for(int i=(b);i>=a;i--)
#define MEM1(a) memset(a,0,sizeof(a))
#define MEM2(a) memset(a,-1,sizeof(a))
const double PI=acos(-1.0);
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
using namespace std;
#define ll __int64
int n,m,t;
#define Mod 1000000007
#define N 110
#define M 1000100
priority_queue<ll,vector<ll>,greater<ll> >q;
int main(){
#ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
//  freopen("out.txt","w",stdout);
#endif
    	int n;
    	while (sf(n)!=EOF){
    		while(!q.empty())
    			q.pop();
    		ll x;
    		for(int i=0;i<n;i++){
    			sfI(x);
    			q.push(x);
    		}
    		ll sum = (q.size()==1)?x:0;
    		while(q.size()>1){
    			ll a = q.top();
    			q.pop();
    			ll b = q.top();
    			q.pop();
    			sum += (a+b);
    			q.push(a+b);
    		}
    		printf("%I64d\n",sum);
    	}
return 0;
}