【poj2431】驾驶问题-贪心,优先队列
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 29360 | Accepted: 8135 |
Description
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
* 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
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
#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
- 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, ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- [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 ...
- 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 ...
- 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/ ...
- 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
- prometheus+grafana监控redis
prometheus+grafana监控redis redis安装配置 https://www.cnblogs.com/autohome7390/p/6433956.html redis_export ...
- Golang语言编程规范
Golang语言编程规范 一.说明 编程规范好,可避免语言陷阱,可有利团队协作,有利项目维护. 正常的Go编程规范有两种:编译器强制的(必须的),gofmt格式化非强制的(非必须). Go宣告支持驼峰 ...
- (七)装配Bean(1)
针对给接口提供哪一个具体的实现,也就是装配哪一种具体的实现bean,在Spring中提供了多种方式,主要包括3种: 一.隐式的bean发现机制和自动装配(自动化装配bean) 二.在Java类中进行显 ...
- element的Dialog组件踩坑
在一个组件页面中需要有一个弹窗,为了代码简洁我把弹窗封装成一个组件方便重复调用 描述大致是一个父组件,里面有一个按钮还有一个子组件(弹窗),点击按钮让弹窗出来,弹窗自带的有关闭功能,点击关闭以后再点击 ...
- 内涵段子——脑筋急转弯——spider
# python 3.7 from urllib.request import Request,urlopen import re,time class Neihan(object): def __i ...
- 初级文件IO——IO过程、open、close、write、read、lseek、dup、dup2、errno、perror
先要回答的问题 文件IO指的是什么? 本文主要讲述如何调用Linux OS所提供的相关的OS API,实现文件的读写. 如何理解文件IO? IO就是input output的意思,文件io就是文件输入 ...
- C# 设置鼠标光标位置
C# 设置鼠标光标位置 using System.Drawing; using System.Runtime.InteropServices; namespace ZB.QueueSys.Common ...
- 如何利用while语句打印“九九乘法口诀表”
需求:输出九九乘法表 plus.py代码如下: i=1 j=1 while i<=9: j=1 while j<=i: print(j,'*',i,'=',str(i*j)+' ',end ...
- Nginx系列1.2:nginx-rtmp流媒体服务器添加权限认证(推流权限和播放权限)
用到的工具:OBS Studio(推流).nginx-rtmp流媒体服务器.VLC(拉取流播放) Nginx系列1:ubuntu16.04编译出适合自己的nginx服务器 Nginx系列1.1:ubu ...
- jade总结
随着时间的迁移,要跟官方api相匹配 jade的缺点 1.可移植性差 2.调试困难 3.性能不是非常出色(不是为性能设计,可以使用dot, http://olado.github.io/) 选择的 ...