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
Lines 2..N+1: Each line contains a single integer describing the length of a needed plank
Output
Sample Input
3 8 5 8
Sample Output
34
Hint
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).
FJ需要修补牧场的围栏，他需要 N 块长度为 Li 的木头（N planks of woods）。开始时，FJ只有一块无限长的木板，
因此他需要把无限长的木板锯成 N 块长度为 Li 的木板，Farmer Don提供FJ锯子，但必须要收费的，收费的标准是对应每次据出木块的长度，
比如说测试数据中 5 8 8，一开始，FJ需要在无限长的木板上锯下长度 21 的木板（5+8+8=21），第二次锯下长度为 5 的木板，
第三次锯下长度为 8 的木板，至此就可以将长度分别为 5 8 8 的木板找出。
题解:
1.贪心
显然，通过每次选取两块长度最短的木板，合并，最终必定可以合并出长度为 Sum（Li）的木板，并且可以保证总的耗费最少，总时间复杂度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 //合并 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.优先队列
思想跟贪心的一样，由于只需从板的集合里取出最短的两块，并且把长度为两块板长度之和的板加入集合中即可，因此使用优先队列可以高效实现。
一共需要O（n）次O（logn）的操作，因此总时间复杂度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<<" "; 24 } 25 return 0; 26 }