贪心+优先队列（哈夫曼思想）POJ 3253 Fence Repair

Fence Repair
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 33173 Accepted: 10714
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
Hint
He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into 16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).
Source
USACO 2006 November Gold

有一个农夫要把一个木板钜成几块给定长度的小木板，每次锯都要收取一定费用，这个费用就是当前锯下的这个木版的长度。给定各个要求的小木板的长度，及小木板的个数n，求最小费用。
可以看成是把n段木板拼成一块，每次拼的代价为所拼木板的长度和。这就跟哈夫曼编码一样，每次选取两个最小的来拼。具体实现时用优先队列。

//876K  47MS#include <stdio.h>#include <queue>#include<iostream>using namespace std;#define INF 0x7fffffffpriority_queue<int,vector<int>,greater<int> > pq;int main(){  freopen("input.txt","r",stdin);  freopen("output.txt","w",stdout);  int n,i,l,a,b;  long long ans;  while(scanf("%d",&n)!=EOF){    for(i=0;i<n;i++){      scanf("%d",&l);      pq.push(l);    }    pq.push(INF);    ans=0;    while(!pq.empty()){      a=pq.top(),pq.pop();      b=pq.top(),pq.pop();      if(b==INF)  break;      ans+=(a+b);    //cout<<" "<<a<<" "<<b<<" "<<ans<<endl;      pq.push(a+b);}printf("%lld\n",ans);  }  return 0;}