Fence Repair
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 37272
Accepted: 12078
Description
Farmer John wants to repair a small length of the fence ar
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 ...
#include
#include
struct cmp
{
bool operator()(const long long &i,const long long &j)
{
return i>j;
}
};
using namespace std;
int main()
{
priority_queuelong long,vectorlong lo
【算法】贪心+堆
#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=20010;
int n,heap[maxn],sz;
void heap_push(int x)
{
heap[++sz]=x;//新数入堆底
int now=sz;...
钢条切割
public class SteelCutting {
public static void main(String[] args) {
SteelCutting s = new SteelCutting();
int[] p = new int[] { 0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 25};
System.out.println
Fence Repair
Time Limit: 2000MS
Memory Limit: 65536K
Total Submissions: 28359
Accepted: 9213
Description
Farmer John wants to repair a small length of the fence aroun...
<em>木板</em>切割<em>问题</em>的优先队列实现方式。
在c++中，首先要定义一个从小到达的优先队列，参见c++代码。
//
// 077_fence repair(pq).cpp
// changlle
//
// Created by user on 12/30/15.
// Copyright (c) 2015 user. All rights reserved.
//
#include
Description
In order to build a ship to travel to Eindhoven, The Netherlands, various sheet metal parts have to be cut from rectangular pieces of sheet metal. Each part is a convex polygon with at mo...
又是一个贪心<em>问题</em>。
这里使用数组遍历来找到第一小与第二小的<em>木板</em>。后续会给出优先队列的实现方式。
//
// 047_fence repair.cpp
// changlle
//
// Created by user on 12/26/15.
// Copyright (c) 2015 user. All rights reserved.
//
#include
#inc
题目连接http://poj.org/problem?id=3253DescriptionFarmer 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,
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 lengt...
/*
poj3252：
农夫为修理栅栏，要将一块很长的<em>木板</em>切成N块，准备切成的<em>木板</em>的长度为L1，L2，L3……Ln。未切割<em>木板</em>之前的长度正好是是要切割的总长度。
每次切断<em>木板</em>时，需要的开销为这块<em>木板</em>的长度，求出按照目标要求将<em>木板</em>切割完成最小的开销是多少。
1<=50000;
*/
# include
# include
# include
using na
