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Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 28155 | Accepted: 9146 |
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
Output
Sample Input
3 8 5 8
Sample Output
34
Hint
Source
题意:有一块很长的木板,要切割成n小块,已知每块的长度。已知分割长为L的木板需要的费用是L,问完成所有切割最少需要多少费用。
解析:Huffman树的思想。我们可以倒过来看,我们不从最大开始分割,而是将分割好的小块一块一块拼回去,结果是一样的。这样的话,无疑就是最优二叉树——Huffman树,所有只要按照Huffman的思想写就行了。
AC代码:
#include <cstdio> #include <iostream> #include <algorithm> #include <queue> using namespace std; int a[20002]; int main(){ #ifdef sxk freopen("in.txt", "r", stdin); #endif //sxk int n, foo; while(scanf("%d", &n)!=EOF){ priority_queue<int, vector<int>, greater<int> > a; for(int i=0; i<n; i++){ scanf("%d", &foo); a.push(foo); } long long ans = 0; //int会WA for(int i=0; i<n-1; i++){ //每次取两个最小的 int x = a.top(); a.pop(); int y = a.top(); a.pop(); ans += x + y; //记录拼接费用 a.push(x + y); //拼接之后,再放回优先级队列中去 } printf("%lld\n", ans); } return 0; }
小编福利:Huffman树的思想还是很巧妙地,很多地方都能用到,没想到A题也能用到,还以为只有数据结构课上学的呢~~~
POJ 3253 Fence Repair (贪心 + Huffman树)
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原文地址:http://blog.csdn.net/u013446688/article/details/43018607