以下代码可对结构体数组中的元素进行排序,也差不多算是一个小小的模板了吧

#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
int x;
int y;
bool operator<(const node &a) const//此操作是对操作符"<"进行重构
{
return x < a.x;//对结构体数组x进行从大到小排序
// return y > a.y;//对结构体数组y进行从大到小排序
}
}s[100]; //bool cmp(int x,int y)//这个重构函数不能用在结构体数组中
//{
// return x > y;
//} int main()
{
int n;
cin >> n;
for(int i = 0;i < n;i++)
cin >> s[i].x >> s[i].y; sort(s,s+n);
for(int i = 0;i < n;i++)
cout << s[i].x << " " << s[i].y << endl;
cout << endl;
return 0;
}

运行结果:

也可以这样

#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
int x;
int y;
bool operator<(const node &a) const
{
return x > a.x;
}
}s[]; bool cmp(node m,node n)
{
m.x < n.x;
} int main()
{
int n;
cin >> n;
for(int i = ;i < n;i++)
cin >> s[i].x >> s[i].y; sort(s,s+n);
for(int i = ;i < n;i++)
cout << s[i].x << " " << s[i].y << endl;
cout << endl;
return ;
}

对优先队列的应用,POJ2431是一个很好的题目,此题用了优先队列+贪心

Expedition
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 29720   Accepted: 8212

Description

A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every unit of distance it travels.

To repair the truck, the cows need to drive to the nearest town (no more than 1,000,000 units distant) down a long, winding road. On this road, between the town and the current location of the truck, there are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire additional fuel (1..100 units at each stop).

The jungle is a dangerous place for humans and is especially dangerous for cows. Therefore, the cows want to make the minimum possible number of stops for fuel on the way to the town. Fortunately, the capacity of the fuel tank on their truck is so large that there is effectively no limit to the amount of fuel it can hold. The truck is currently L units away from the town and has P units of fuel (1 <= P <= 1,000,000).

Determine the minimum number of stops needed to reach the town, or if the cows cannot reach the town at all.

Input

* Line 1: A single integer, N

* Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the town to the stop; the second is the amount of fuel available at that stop.

* Line N+2: Two space-separated integers, L and P

Output

* Line 1: A single integer giving the minimum number of fuel stops necessary to reach the town. If it is not possible to reach the town, output -1.

Sample Input

4
4 4
5 2
11 5
15 10
25 10

Sample Output

2

Hint

INPUT DETAILS:

The truck is 25 units away from the town; the truck has 10 units of fuel. Along the road, there are 4 fuel stops at distances 4, 5, 11, and 15 from the town (so these are initially at distances 21, 20, 14, and 10 from the truck). These fuel stops can supply up to 4, 2, 5, and 10 units of fuel, respectively.

OUTPUT DETAILS:

Drive 10 units, stop to acquire 10 more units of fuel, drive 4 more units, stop to acquire 5 more units of fuel, then drive to the town.

Source

思路:贪心思路为,在假设汽车一路向前,在途中有经过加油站的话,把加油站进行排序,符合条件的加油站中的汽油从大到小排序,存到优先队列中,在汽车用完时,即随时可以在队列的队头取汽油,使得汽车尽可能加最多的汽油,停最少次
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
struct node
{
int dis;
int fuel;
bool operator<(const node &a) const//此重构操作,请参考上面的代码
{
return dis > a.dis;//返回dis,说明是对dis这个数组进行排序操作
}
}stop[]; priority_queue<int> que; int main()
{
int n,l,p;
cin >> n;
for(int i = ;i < n;i++)
cin >> stop[i].dis >> stop[i].fuel;
cin >> l >> p;
int ans = ;
sort(stop,stop + n);//对加油站距离dis数组进行从大到小排序
que.push(p);//队列自动从大到小排序,即排序汽油p
int temp = ;
while(l > && !que.empty())//卡车未到达终点并且卡车在当前汽油用完前有路过加油站
{
ans++;//卡车停下加一次油计数器
l -= que.top();//加油,更新一次汽油用尽后距离终点的距离
que.pop();//删除已用的加油站的汽油
while(l <= stop[temp].dis && temp < n)//卡车距离终点的距离小于等于最近加油站的距离并且这个加油站的位置在终点加油站前面,这里假设终点也为一个加油站。//l <= stop[i].dis意思是卡车能经过离它最近的一个加油站,如果大于的话,说明卡车停下时没有加油站可加油 
que.push(stop[temp++].fuel);//将经过的加油站压入优先队列,要使用的时候就取队头元素(队头中存的汽油最大) //如果经过加油站,则一定要将该加油站的可加油量添加到优先队列当中 
}//temp++说明离卡车最近的加油站,卡车继续往前开,加油站点也依次往后,所以变量temp需要自增
if(l > )//如果卡车距离终点的距离还大于0的话,即通过不了终点
cout << "-1" << endl;
else
cout << ans - << endl;//在起点深度时候记为一次加油,这里需要减去1
return ;
}

其中代码有详细的注释,希望注释能加深理解

代码参考于:博客园-小小菜鸟

POJ2431 优先队列+贪心 - biaobiao88的更多相关文章

  1. 最高的奖励 - 优先队列&贪心 / 并查集

    题目地址:http://www.51cpc.com/web/problem.php?id=1587 Summarize: 优先队列&贪心: 1. 按价值最高排序,价值相同则按完成时间越晚为先: ...

  2. hdu3438 Buy and Resell(优先队列+贪心)

    Buy and Resell Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)To ...

  3. poj2431(优先队列+贪心)

    题目链接:http://poj.org/problem?id=2431 题目大意:一辆卡车,初始时,距离终点L,油量为P,在起点到终点途中有n个加油站,每个加油站油量有限,而卡车的油箱容量无限,卡车在 ...

  4. H - Expedition 优先队列 贪心

    来源poj2431 A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being ...

  5. ZOJ-3410Layton's Escape(优先队列+贪心)

    Layton's Escape Time Limit: 2 Seconds      Memory Limit: 65536 KB Professor Layton is a renowned arc ...

  6. CodeForces - 853A Planning (优先队列,贪心)

    Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n ...

  7. 1350: To Add Which? (优先队列+贪心 或者 数组模拟)

    1350: To Add Which? Submit Page    Summary    Time Limit: 1 Sec     Memory Limit: 128 Mb     Submitt ...

  8. 牛客网 桂林电子科技大学第三届ACM程序设计竞赛 C.二元-K个二元组最小值和最大-优先队列+贪心(思维)

    链接:https://ac.nowcoder.com/acm/contest/558/C来源:牛客网 小猫在研究二元组. 小猫在研究最大值. 给定N个二元组(a1,b1),(a2,b2),…,(aN, ...

  9. hdu 4544 优先队列+贪心

    题意:最近,减肥失败的湫湫为发泄心中郁闷,在玩一个消灭免子的游戏.游戏规则很简单,用箭杀死免子即可.箭是一种消耗品,已知有M种不同类型的箭可以选择,并且每种箭都会对兔子造成伤害,对应的伤害值分别为Di ...

随机推荐

  1. ZooKeeper单机服务端的启动源码阅读

    程序的入口QuorumPeerMain public static void main(String[] args) { // QuorumPeerMain main = new QuorumPeer ...

  2. springboot系列之04-提高开发效率必备工具lombok

    未经允许,不得转载 原作者:字母哥博客 本文完整系列出自:springboot深入浅出系列 一.前置说明 本节大纲 使用lombok插件的好处 如何安装lombok插件 使用lombok提高开发效率 ...

  3. 利用百度云接口实现车牌识别·python

    一个小需求---实现车牌识别. 目前有两个想法 1. 调云在线的接口或者使用SDK做开发(配置环境和变异第三方库麻烦,当然使用python可以避免这些问题) 2. 自己实现车牌识别算法(复杂) 一开始 ...

  4. [一]基本sqlplus命令

    基本sqlplus命令: 1: sqlplus scott/tiger ; #简化连接数据库 2:show user; #想知道当前登陆的用户是哪一位 3:conn 用户名[/密码] [AS SYSD ...

  5. 快学Scala 第一课 (变量,类型,操作符)

    Scala 用val定义常量,用var定义变量. 常量重新赋值就会报错. 变量没有问题. 注意:我们不需要给出值或者变量的类型,scala初始化表达式会自己推断出来.当然我们也可以指定类型. 多个值和 ...

  6. java中List、Set和Map三个接口及其主要实现类

    三个接口都在java.util包下 List与Set具有相似性,它们都是单列元素的集合,所以,它们有一个共同的父接口,叫Collection,Map没有继承Collection接口 1.List接口: ...

  7. day 19作业

    目录 今日作业: 今日作业: 1.什么是对象?什么是类? 答:对象是特征与技能的集合体,类是一系列对象相同的特征与技能的结合体 2.绑定方法的有什么特点 答:由对象来调用称之为对象的绑定方法,不同的对 ...

  8. 在chrome浏览器中调用IE浏览器并访问(openIE.reg自定义协议)

    在谷歌浏览器中有4种方法调用IE浏览器.如下: c++ socket通过浏览器在ie中打开指定url : vb生成exe,url访问exe启动ie并打开指定url : 通过socket实现通过http ...

  9. Travis CI持续集成使用

    用好这个工具不仅可以提高效率,还能使开发流程更可靠和专业化,从而提高软件的价值.而且,它对于开源项目是免费的,不花一分钱,就能帮你做掉很多事情. 一.什么是持续集成? Travis CI 提供的是持续 ...

  10. Jenkins构建Jmeter项目之源代码管理(SVN)

    1.查看项目创建中是否又svn插件,没有的话下载插件subversion 2.配置svn源代码管理,如下图(testcases目录下包含build.xml和脚本文件) 3.查看Jenkins本地工作空 ...