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 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.
问你要到达目的地,最少需要下几次车,用贪心的思想每次只拿最大的,先对加油站的位置排个序,然后现在油量能到的加油站就push进优先队列,按油量大到小排,一次加一个,到了就跳出,队列空了就到不了
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define prf(x) printf("%d\n",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
const ll mod=1e9+7;
const double eps=1e-8;
const int inf=0x3f3f3f3f;
using namespace std;
const double pi=acos(-1.0);
const int N=1e4+10;
struct node
{
int pos,fue;
}a[N];
priority_queue<int>v;
bool cmp(node a,node b)
{
return a.pos>b.pos;
}
int main()
{
int ans=0,n,l,now;scf(n);
rep(i,0,n)
scff(a[i].pos,a[i].fue);
scff(l,now);
sort(a,a+n,cmp);
int i=0;
while(now<l)
{
for(;i<n;i++)
{
if(l-a[i].pos>now) break;
v.push(a[i].fue);
}
if(v.empty())
{
pf("-1\n");return 0;
}
now+=v.top();
v.pop();
ans++;
}
prf(ans);
return 0;
}
H - Expedition 优先队列 贪心的更多相关文章
- poj 3431 Expedition 优先队列
poj 3431 Expedition 优先队列 题目链接: http://poj.org/problem?id=2431 思路: 优先队列.对于一段能够达到的距离,优先选择其中能够加油最多的站点,这 ...
- POJ2431 优先队列+贪心 - biaobiao88
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 namespa ...
- 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 ...
- POJ 2431 Expedition【贪心】
题意: 卡车每走一个单元消耗一升汽油,中途有加油站,可以进行加油,问能否到达终点,求最少加油次数. 分析: 优先队列+贪心 代码: #include<iostream> #include& ...
- 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: ...
- POJ 2431 Expedition (优先队列+贪心)
题目链接 Description A group of cows grabbed a truck and ventured on an expedition deep into the jungle. ...
- POJ 2431 Expedition (贪心 + 优先队列)
题目链接:http://poj.org/problem?id=2431 题意:一辆卡车要行驶L单位距离,卡车上有P单位的汽油.一共有N个加油站,分别给出加油站距终点距离,及加油站可以加的油量.问卡车能 ...
- POJ 2431——Expedition(贪心,优先队列)
链接:http://poj.org/problem?id=2431 题解 #include<iostream> #include<algorithm> #include< ...
- 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 ...
随机推荐
- python之编程风格
第一:语句和语法 # 表示注释掉的内容 \ 续行 print("yyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy\ yyyyyyyyyyyyyyyyyyyyy ...
- 如何移除HTML5 input在type="number"时的上下小箭头
在chrome下: input::-webkit-outer-spin-button,input::-webkit-inner-spin-button{ -webkit-appearance ...
- Coursera机器学习+deeplearning.ai+斯坦福CS231n
日志 20170410 Coursera机器学习 2017.11.28 update deeplearning 台大的机器学习课程:台湾大学林轩田和李宏毅机器学习课程 Coursera机器学习 Wee ...
- method.invoke(...)反射点
import java.lang.reflect.Method; import java.util.Arrays; /** * @Author: hoobey * @Description: * @D ...
- PL/SQL学习笔记之日期时间
一:PL/SQL时间相关类型 PL/SQL提供两个和日期时间相关的数据类型: 日期时间(Datetime)数据类型 时间间隔类型 二:日期时间类型 datetime数据类型有: DATE TIMEST ...
- 【Docker江湖】之docker部署与理解
转载请注明出处:http://blog.csdn.net/gamer_gyt 博主微博:http://weibo.com/234654758 Github:https://github.com/thi ...
- Pinterest凭什么拥有那么多用户:机器学习是答案
目前,Pinterest月平均活跃用户量达到1亿,这家以图片为主的公司是如何留住用户并盈利的呢?Pinterest的主要目标是向用户推荐相关的图片或内容,推荐的内容足够精确才能提高用户黏性.近期,&l ...
- keil软件错误总结.doc
KEIL编译错误信息表 错误代码及错误信息 错误释义 error 1: Out of memory 内存溢出 error 2: Identifier expected 缺标识符 error 3: ...
- 【ThinkPHP】ThinkPHP环境的安装与配置
ThinkPHP是一个免费开源的,快速.简单的面向对象的轻量级PHP开发框架. 严格来说,ThinkPHP无需安装过程,这里所说的安装其实就是把ThinkPHP框架放入WEB运行环境(前提是你的WEB ...
- word,excel,ppt,txt转换为 PDF
/// <summary> /// 将word文档转换成PDF格式 /// </summary> /// <param name="sourcePath&quo ...