POJ3253 Fence Repair 贪心+优先队列(堆维护)
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1.题目描述:
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
3858
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).
Source
2.题意概述:
一块长木板,要经过n-1次切割将其切成n块FJ想要的木板,对于每块木板,没切割一次,将会消耗和这条木板长度值相等的金钱,问最少需要多少钱,可将木板切成自己想要的n块。
3.解题思路:
类似于合并果子,现在想想,应该解题思路会有很多,dp对于这个数据肯定会超时,然后最开始poj做那道题就是二叉树+贪心结果狠狠地超时,当时集训做的时候就是用的堆维护,NlogN确实棒
4.AC代码:
#include <iostream>#include <algorithm>#include <functional>#include <queue>#include <cstring>#include<cstdio>using namespace std;typedef long long ll;ll sum;int n, t, a, b;int main(){while (scanf("%d", &n) != EOF){priority_queue<int, vector<int>, greater<int> >q;for (int i = 0; i < n; i++){scanf("%d", &t);q.push(t);}sum = 0;if (q.size() == 1){a = q.top();sum += a;q.pop();}while (q.size() > 1){a = q.top();q.pop();b = q.top();q.pop();t = a + b;sum += t;q.push(t);}printf("%lld\n", sum);}return 0;}
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