POJ-5353-Fence Repair(哈夫曼问题->贪心（一种解法两种做法）)



Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 40578		Accepted: 13273

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).
题意：给一块长木板，现要将其锯成n段，共需锯n-1次，每次锯的代价为所锯木板的长度，求最小总代价。

题解：
利用Huffman思想，要使总费用最小，那么每次只选取最小长度的两块木板相加，再把这些“和”累加到总费用中即可

本题虽然利用了Huffman思想，但是直接用HuffmanTree做会超时，可以用优先队列做

做法一：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<limits.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<string>
#include<math.h>
#include<map>
using namespace std;
#define inf 1e18
#define maxn 20005
long long a[maxn];
long long n;
long long ans;
void Swap(int a,int b)
{
int temp;
temp=a;a=b;b=temp;
}
void solve()
{
ans=0;
while(n>1)//直到计算到木板为1块时为止
{
//求出最短的板x和次短的板y
int x=0,y=1;
if(a[x]>a[y])
Swap(x,y);
for(int i=2;i<n;i++)
{
if(a[i]<a[x])
{
y=x;
x=i;
}
if(a[i]<=a[y])
y=i;
}
long long temp=a[x];//保存最小值
a[x]=inf;
for(int i=0;i<n;i++)//找寻次小值
{
if(a[i]<a[y])
y=i;
}
//将两板结合
a[x]=temp;
long long t=a[x]+a[y];
ans+=t;
if(x==n-1)
Swap(x,y);
a[x]=t;
a[y]=a[n-1];
n--;
}
printf("%lld\n",ans);
}
int  main()
{
int i,j;
while(scanf("%lld",&n)!=EOF)
{
memset(a,0,sizeof(a));
for(i=0;i<n;i++)
scanf("%lld",&a[i]);
solve();
}
}

做法二：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
#include<limits.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<vector>
#include<string>
#include<math.h>
#include<map>
using namespace std;
#define maxn 20005
priority_queue<int,vector<int>,greater<int> >q;
int a[maxn];
int main()
{
int i,j,n,x,y;
long long ans=0;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
q.push(a[i]);
}
while(q.size()>1)
{
x=q.top();
q.pop();
y=q.top();
q.pop();
int t=x+y;
ans+=t;
q.push(t);
}
printf("%lld\n",ans);
}