Fence Repair
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 66577 | Accepted: 21929 |
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
这一题其实是运用了霍夫曼编码的原则，简单说就是将用到次数最少的数放在树的最底层，用到次数最多的放在最高层，如果想要多了解可以百度搜一下霍夫曼编码
AC源码如下：
#include<iostream>
#include<algorithm>
using namespace std;
int L[100000],n;
int main()
{
cin>>n;
for(int i=0;i<n;i++){
cin>>L[i];
}
long long ans=0;
while(n>1){
int min1=0,min2=1;
if(L[min1] > L[min2])swap(min1,min2);
for(int i = 2;i < n;i++){
if(L[i]<L[min1]){
min2=min1;
min1=i;
}else if(L[i] < L[min2]){
min2=i;
}
}
int t=L[min1]+L[min2];
ans+=t;
L[min1]=t;
L[min2]=L[n-1];
n--;
}
cout<<ans<<endl;
return 0;
}