Expedition
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 29360   Accepted: 8135

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:
  一直走,把能到的加油站扔到大根堆里,每次取出油量最大的加油站加油,一直到终点,记录次数。
代码:
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<queue>
#include<algorithm>
#define int long long
#define N 100050
using namespace std;
int n,L,p,sum,ans,cnt;
struct node
{
int dis,l;
bool operator < (const node &a) const
{
return dis > a.dis;
}
}e[N];
priority_queue <int> q;
signed main()
{
scanf("%lld",&n);
for(int i=0,a,b;i<n;i++)
{
scanf("%lld%lld",&e[i].dis,&e[i].l);
}
scanf("%lld%lld",&L,&p);
int tmp=0;
q.push(p);
sort(e,e+n);
while(L>0&&!q.empty())
{
L-=q.top();
ans++;
q.pop();
while(L<=e[tmp].dis&&tmp<n)
q.push(e[tmp++].l);
}
if(L>0) printf("-1\n");
else printf("%d\n",ans-1);
return 0;
}

【poj2431】驾驶问题-贪心,优先队列的更多相关文章

  1. hihoCoder 1309:任务分配 贪心 优先队列

    #1309 : 任务分配 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 给定 N 项任务的起至时间( S1, E1 ), ( S2, E2 ), ..., ( SN,  ...

  2. UVA 11134 - Fabled Rooks(贪心+优先队列)

    We would like to place  n  rooks, 1 ≤  n  ≤ 5000, on a  n×n  board subject to the following restrict ...

  3. C. Playlist Educational Codeforces Round 62 (Rated for Div. 2) 贪心+优先队列

    C. Playlist time limit per test 2 seconds memory limit per test 256 megabytes input standard input o ...

  4. HDU 6438 网络赛 Buy and Resell(贪心 + 优先队列)题解

    思路:维护一个递增队列,如果当天的w比队首大,那么我们给收益增加 w - q.top(),这里的意思可以理解为w对总收益的贡献而不是真正获利的具体数额,这样我们就能求出最大收益.注意一下,如果w对收益 ...

  5. 贪心+优先队列 HDOJ 5360 Hiking

    题目传送门 /* 题意:求邀请顺序使得去爬山的人最多,每个人有去的条件 贪心+优先队列:首先按照l和r从小到大排序,每一次将当前人数相同的被邀请者入队,那么只要能当前人数比最多人数条件小,该人能 被邀 ...

  6. [POJ1456]Supermarket(贪心 + 优先队列 || 并查集)

    传送门 1.贪心 + 优先队列 按照时间排序从前往后 很简单不多说 ——代码 #include <queue> #include <cstdio> #include <i ...

  7. Painting The Fence(贪心+优先队列)

    Painting The Fence(贪心+优先队列) 题目大意:给 m 种数字,一共 n 个,从前往后填,相同的数字最多 k 个在一起,输出构造方案,没有则输出"-1". 解题思 ...

  8. CF140C New Year Snowmen(贪心+优先队列)

    CF140C 贪心+优先队列 贪心策略:每次取出数量最多的三种球,合成一个答案,再把雪球数都-1再插回去,只要还剩下三种雪球就可以不断地合成 雪球数用优先队列维护 #include <bits/ ...

  9. BZOJ1029: [JSOI2007]建筑抢修[模拟 贪心 优先队列]

    1029: [JSOI2007]建筑抢修 Time Limit: 4 Sec  Memory Limit: 162 MBSubmit: 3785  Solved: 1747[Submit][Statu ...

随机推荐

  1. django静态文件配置和使用

    一.首先需要了解的知识点是: 1.出于对效率和安全的考虑,django管理静态文件的功能仅限于在开发阶段的debug模式下使用,且需要在配置文件的INSTALLED_APPS中加入django.con ...

  2. (十)SpringBoot之web 应用开发-Servlets, Filters, listeners

    一.需求 Web 开发使用 Controller 基本上可以完成大部分需求,但是我们还可能会用到 Servlet. FilterListene 二.案例 2.1 通过注册 ServletRegistr ...

  3. C#获取Excel表格所有sheet名(Epplus)

    原文:C#获取Excel表格所有sheet名(Epplus) 版权声明:本文为博主原创文章,遵循CC 4.0 BY-SA版权协议,转载请附上原文出处链接和本声明. 本文链接:https://blog. ...

  4. Go 缓冲信道

    缓冲信道 语法结构:cap为容量 ch := make(chan type, cap) 缓冲信道支持len()和cap(). 只能向缓冲信道发送容量以内的数据. 只能接收缓冲信道长度以内的数据. 缓冲 ...

  5. iOS - 解决警告“ld: Warning: Directory Not Found for Option”

    有时候我们可能从项目中删除了某个目录.文件以后,编译出现警告信息: ld: warning: directory not found for option“XXXXXX” 具体类似下图: 很奇怪,为什 ...

  6. 如何使用Visual Studio Code调试PHP CLI应用和Web应用

    在按照Jerry的公众号文章 什么?在SAP中国研究院里还需要会PHP开发? 进行XDebug在本地的配置之后,如果想使用Visual Studio Code而不是Eclipse来调试PHP应用,步骤 ...

  7. CentOS 7安装Hadoop集群

    准备三台虚拟机,ip分别为192.168.220.10(master).192.168.220.11(slave1).192.168.220.12(slave2) 准备好jdk-6u45-linux- ...

  8. 网络流dinic ek模板 poj1273

    这里只是用来存放模板,几乎没有讲解,要看讲解网上应该很多吧…… ek bfs不停寻找增广路到找不到为止,找到终点时用pre回溯,O(VE^2) #include<cstdio> #incl ...

  9. Free lunch is over

    译文:http://www.mamicode.com/info-detail-1324737.html 原文:http://www.gotw.ca/publications/concurrency-d ...

  10. docker安装redis并以配置文件方式启动

    镜像相关 redis镜像 # 不限定版本 docker pull redis # 拉取 redis为4.0.9版本的镜像 docker pull redis:4.0.9 # 拉取之后查看镜像 dock ...