解题报告：
Fence Repair
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Time Limit: 2000MS |
Memory Limit: 65536K |
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Total Submissions: 39958 |
Accepted: 13033 |
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
题意：
要8,5,8长的木板，最少要截多少钱，首先是21米的，要截21元，截成了8,13，再截13，要13元截成了5,8，求最少要多少钱。
分析：
刚一开始，感觉是贪心，每次截出最大的，这样就变成了不断求部分和，然而，数据量还算比较大，又想到线段树。WA了好几次，才想到是方案不对。
哈夫曼编码，首先选出最小的两个，截他们的费用最少，但是在此之前，这两节又是原先的木板中截出来的，这样就想到了是一个哈夫曼编码的原理。
#include <stdio.h> #include <queue> using namespace std; int main() { int n; scanf("%d",&n); priority_queue<int,vector<int>,greater<int> > Q; int tmp; for(int i=0;i<n;i++) { scanf("%d",&tmp); Q.push(tmp); } long long sum =0; while(Q.size()>1) { int a = Q.top(); Q.pop(); int b = Q.top(); Q.pop(); sum+=(a+b); Q.push(a+b); } printf("%lld\n",sum); return 0; }