POJ 3253 Fence Repair (贪心)

Fence Repair

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 3253

Appoint description: 
hanjiangtao  (2014-11-12)
System Crawler  (2015-04-24)

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

反过来考虑，将一堆木板组合成一个，每次选最小的两个，用个优先队列来维护。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <cctype>
 #include <cmath>
 #include <queue>
 #include <map>
 #include <cstdlib>
 #include <vector>
 using    namespace    std;

 int    n;
 int    box;
 long    long    a,b;
 long    long    sum;
 priority_queue<long long,vector<long long>,greater<long long> >    que;

 int    main(void)
 {
     while(scanf("%d",&n) != EOF)
     {
         for(int i = ;i < n;i ++)
         {
             scanf("%d",&box);
             que.push(box);
         }

         sum = ;
         while(que.size() != )
         {
             a = que.top();
             que.pop();
             sum += a;
             b = que.top();
             que.pop();
             sum += b;
             que.push(a + b);
         }
         que.pop();
         printf("%lld\n",sum);
     }

     return    ;
 }