POJ 3253

贪心思路：参考挑战书

简单说就是最小的两个木板应该是同一个大木板切出来的，以此类推，我用的是优先队列，感觉比书上的简单好理解。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
#define fo freopen("in.txt","r",stdin)
#define fc fclose(stdin)
#define fu0(i,n) for(i=0;i<n;i++)
#define fu1(i,n) for(i=1;i<=n;i++)
#define fd0(i,n) for(i=n-1;i>=0;i--)
#define fd1(i,n) for(i=n;i>0;i--)
#define mst(a,b) memset(a,b,sizeof(a))
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d %d",&n,&m)
#define ss(s) scanf("%s",s)
#define sddd(n,m,k) scanf("%d %d %d",&n,&m,&k)
#define pans(ans) printf("%d\n",ans)
#define all(a) a.begin(),a.end()
#define sc(c) scanf("%c",&c)
const int maxn=200005;
const double eps=1e-8;
const double PI = acos(-1.0);
int main()
{
    int n,i;
    while(cin>>n)
    {
        int l[20005];
        priority_queue <int,vector<int>,greater<int> > pq;
        fu0(i,n)
        {
            sd(l[i]);
            pq.push(l[i]);
        }
        ll ans=0;
        while(n>1)//最后合并为1个木块
        {
            int tp=pq.top();
            pq.pop();
            int tpp=pq.top();
            pq.pop();
            int t=tp+tpp;//合并
            ans+=t;
            pq.push(t);
            n--;
        }
        cout<<ans<<endl;
    }
    return 0;
}