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 Nplanks (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 plankOutput
Line 1: One integer: the minimum amount of money he must spend to make N-1 cutsSample Input
3 8 5 8Sample Output
34
一道贪心题。题意是要把木板锯成若干段，每次锯木板的代价是锯后两段木板之和。根据贪心思想，由逆向思维，每次选取最短的两段即可。用优先队列即可。（
#include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <cmath> #include <cstring> #include <queue> #include <string> #define N 1e5+10 using namespace std; struct cmp { bool operator()(int x,int y) { return x>y; } }; priority_queue<int,vector<int>,cmp> Q; int main() { int n,t; cin>>n; long long ans=0; if(n==1) { cout<<n<<endl; return 0; } for(int i=0;i<n;i++) { cin>>t; Q.push(t); } while(Q.size()>=2) { int a=Q.top(); Q.pop(); int b=Q.top(); Q.pop(); ans+=(a+b); Q.push(a+b); } cout<<ans<<endl; } 《挑战》上说直接贪心写或者Huffman树，这个现在不想写 留着以后可补）
