poj3253专题

poj3253 FenceRepair

大致题意: 有一个农夫要把一个木板钜成几块给定长度的小木板,每次锯都要收取一定费用,这个费用就是当前锯的这个木版的长度 给定各个要求的小木板的长度,及小木板的个数n,求最小费用   提示: 以 3 5 8 5为例: 先从无限长的木板上锯下长度为 21 的木板,花费 21 再从长度为21的木板上锯下长度为5的木板,花费5 再从长度为16的木板上锯下长度为8的

poj3253 Fence Repair

 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

POJ3253题解(哈夫曼树)

#include<iostream>#include<queue>#include<vector>using namespace std;int main(){int n;cin>>n;priority_queue<int,vector<int>,greater<int>>q;//优先队列,从小到大排序。for(int i=0;i<n;i++){int temp;cin>>temp;q.

poj3253(优先队列 哈夫曼树)

Fence Repair Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 39511 Accepted: 12864 Description Farmer John wants to repair a small length of the fence around the pasture(草地). He mea