contain tput pst pie output 優先 lines names -s

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 L_i (1 ≤ L_i ≤ 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 L_i

). 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). 題意：

FJ需要修補牧場的圍欄，他需要 N 塊長度為 Li 的木頭（N planks of woods）。開始時，FJ只有一塊無限長的木板，
因此他需要把無限長的木板鋸成 N 塊長度為 Li 的木板，Farmer Don提供FJ鋸子，但必須要收費的，收費的標準是對應每次據出木塊的長度，
比如說測試數據中 5 8 8，一開始，FJ需要在無限長的木板上鋸下長度 21 的木板（5+8+8=21），第二次鋸下長度為 5 的木板，
第三次鋸下長度為 8 的木板，至此就可以將長度分別為 5 8 8 的木板找出。

題解:
1.貪心

　　顯然，通過每次選取兩塊長度最短的木板，合並，最終必定可以合並出長度為 Sum（Li）的木板，並且可以保證總的耗費最少，總時間復雜度O（n^2）。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #define ll long long
 6 using namespace std;
 7 int n,l[20005];
 8 int main(){
 9     scanf("%d",&n);for(int i=0;i<n;i++)scanf("%d",&l[i]);
10     ll ans=0;
11     while(n>1){
12         int min1=0,min2=1;  //最短板和次短板 
13         if(l[min1]>l[min2]) swap(min1,min2);
14         for(int i=2;i<n;i++){
15             if(l[i]<l[min1]){
16                 min2=min1;
17                 min1=i;
18             }
19             else if(l[i]<l[min2]) min2=i;
20         }
21         //合並 
22         int t=l[min1]+l[min2];
23         ans+=t;
24         if(min1==n-1) swap(min1,min2);
25         l[min1]=t;
26         l[min2]=l[n-1];
27         n--;
28     }
29     printf("%lld",ans);
30     return 0;
31 }

2.優先隊列　　
　思想跟貪心的一樣，由於只需從板的集合裏取出最短的兩塊，並且把長度為兩塊板長度之和的板加入集合中即可，因此使用優先隊列可以高效實現。　　
　一共需要O（n）次O（logn）的操作，因此總時間復雜度O（nlogn）。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #define maxn 20005
 7 #define ll long long
 8 using namespace std;
 9 int n,a[maxn];
10 int main(){
11     //ios::sync_with_stdio(false);
12     while(~scanf("%d",&n)){
13         for(int i=1;i<=n;i++) cin>>a[i];
14         ll ans=0;
15         priority_queue<int ,vector<int> ,greater<int> > q;
16         for(int i=1;i<=n;i++) q.push(a[i]);
17         while(q.size()>1){
18             int l1=q.top();q.pop();
19             int l2=q.top();q.pop();
20             ans+=(l1+l2);
21             q.push(l1+l2);
22         }
23         cout<<ans<<"\n";
24     }
25     return 0; 
26 }

【poj 3253】Fence Repair