poj3253——Fence Repair（哈夫曼）

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length Li (1 ≤ Li ≤ 50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths Li). FJ is ignoring the “kerf”, the extra length lost to sawdust when a sawcut is made; you should ignore it, too.

FJ sadly realizes that he doesn’t own a saw with which to cut the wood, so he mosies over to Farmer Don’s Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn’t lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer N, the number of planks
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output

Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts
Sample Input

3
8
5
8
Sample Output

34

有一块无限长的木板，从中锯下长度为L的木板就要收取L费用。样例中一个人要长度为8,5,8的三块木板，那就最少需要收取的费用为34，因为先是21=8+8+5，然后是8，最后是5。反过来就是求一个哈夫曼树

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <map>#include <cmath>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 205#define Mod 10001using namespace std;int main(){    int n,x;    long long sum;    while(~scanf("%d",&n))    {        priority_queue<int,vector<int>,greater<int> > q;        if(n==1)        {            scanf("%d",&x);            printf("%d\n",x);        }        else        {            sum=0;            while(n--)            {                scanf("%d",&x);                q.push(x);            }            while(q.size()>1)            {                int a=q.top();                q.pop();                int b=q.top();                q.pop();                int t=a+b;                sum+=t;                q.push(t);            }            printf("%I64d\n",sum);        }    }    return 0;}