Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 67319   Accepted: 22142 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)…

poj 3253 Fence Repair 优先队列 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) u…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 51411   Accepted: 16879 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)…

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 ≤ Li ≤ 50,000) units. He then purchases a…

Fence Repair Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 51346   Accepted: 16857 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)…

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 s…

题目描述: 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…

http://poj.org/problem?id=3253 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…

分割木板的顺序是自由的,所以每次选择两块最短的板,组合在一起,增加队列,原来两个板出队,直到队列中为空或者仅仅剩下一个板时结束.这里使用优先队列较为方便. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #define ll __int64 using namespace std; int len[20005]; i…

题目链接:http://poj.org/problem?id=3253 思路分析:题目与哈夫曼编码原理相同,使用优先队列与贪心思想:读入数据在优先队列中,弹出两个数计算它们的和,再压入队列中: 代码如下: #include <iostream> #include <queue> using namespace std; struct cmp { bool operator() (long long a, long long b) { return a > b; } }; in…