Expedition---POJ - 2431
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
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.
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
const int maxn =1e5+;
int a[maxn],b[maxn];
int n,l,p; struct node{
int x,y;
}pp[maxn]; bool cmp(struct node a,struct node b){
return a.x<b.x;
} int main(){
cin>>n;
for( int i=; i<n; i++ ){
scanf("%d%d",&pp[i].x,&pp[i].y);
}
cin>>l>>p;
for( int i=; i<n; i++ ){
pp[i].x=l-pp[i].x;
}
sort(pp,pp+n,cmp); pp[n].x=l;
pp[n].y=; // for(int i=0; i<=n; i++){
// printf("%d ",pp[i].x);
// }
// cout<<endl;
// for(int i=0; i<=n; i++ ){
// printf("%d ",pp[i].y);
// }
// cout<<endl; priority_queue<int> que;
int ans=;//加油次数
int pos=;//当前位置
int tank=p;//汽油剩余量
b[n]=;
for(int i=; i<=n; i++ ){
int d=pp[i].x-pos;
// cout<<"d="<<d<<endl;
while(tank-d<){
if(que.empty()){
printf("-1\n");
return ;
}
tank+=que.top();
que.pop();
ans++;
// cout<<"+1"<<endl;
}
// cout<<"qian:"<<tank<<endl;
tank-=d;
// cout<<"hou:"<<tank<<endl;
pos=pp[i].x;
que.push(pp[i].y);
}
printf("%d\n",ans); return ;
}
Expedition---POJ - 2431的更多相关文章
- Heap:Expedition(POJ 2431)
远征队 题目大意:一部车要从一个地方走到另一个地方,开始的时候车的油箱有P升油,汽车每走1个距离消耗1升油,没有油汽车无法行驶,路上有加油站,可以为汽车加油,设汽车的油缸是无限大小的,问你汽车能否走到 ...
- POJ 2431 Expedition(探险)
POJ 2431 Expedition(探险) Time Limit: 1000MS Memory Limit: 65536K [Description] [题目描述] A group of co ...
- POJ 2431 Expedition (贪心+优先队列)
题目地址:POJ 2431 将路过的加油站的加油量放到一个优先队列里,每次当油量不够时,就一直加队列里油量最大的直到能够到达下一站为止. 代码例如以下: #include <iostream&g ...
- poj 2431 【优先队列】
poj 2431 Description A group of cows grabbed a truck and ventured on an expedition deep into the jun ...
- POJ 2431 优先队列
汽车每过一单位消耗一单位油,其中有给定加油站可加油,问到达终点加油的最小次数. 做法很多的题,其中优先对列解这题是很经典的想法,枚举每个加油站,判断下当前油量是否小于0,小于0就在前面挑最大几个直至油 ...
- POJ 2431 Expedition (STL 优先权队列)
Expedition Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8053 Accepted: 2359 Descri ...
- poj - 2431 Expedition (优先队列)
http://poj.org/problem?id=2431 你需要驾驶一辆卡车做一次长途旅行,但是卡车每走一单位就会消耗掉一单位的油,如果没有油就走不了,为了修复卡车,卡车需要被开到距离最近的城镇, ...
- 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< ...
- poj 2431 Expedition 贪心 优先队列 题解《挑战程序设计竞赛》
地址 http://poj.org/problem?id=2431 题解 朴素想法就是dfs 经过该点的时候决定是否加油 中间加了一点剪枝 如果加油次数已经比已知最少的加油次数要大或者等于了 那么就剪 ...
随机推荐
- 一, Python 一次性多行打印多个变量
>>> n = 123 >>> f = 456.789 >>> s1 = 'hello ,world' >>> s2 = 'he ...
- 使用element-ui遇到的各种小问题
一.Dialog对话框 1.在使用嵌套Dialog的时候,会出现遮罩层在内容的上方这种错乱情况 解决办法:http://element-cn.eleme.io/#/zh-CN/component/di ...
- JS中定义对象和集合
在js中定义对象: 方式一: var obj = {}; obj['a']=1; obj['b']=2; 方式二: var obj=new Object(); obj.a=1; obj.b=2; 在j ...
- Scyther
1.Security Protocol :a domain analysis 一个安全协议描述了很多的行为,每一个行为称为角色,例如触发角色和 接受角色,一个系统有多个通信代理组成,每一个代理扮演 ...
- 3. ORACLE DATAGUARD 进程
欢迎指正与讨论. 3.1 主库 LNS LNS:一般理解为log network serviceLNS 进程负责将主库redo传输到备库.在11gR1及之前版本进程命名为LNSn,其负责ASYNC和S ...
- Dapper+SqlServerCe部署
C:\Program Files\Microsoft SQL Server Compact Edition\v4.0 目录下面 选择下面这 7 个 SQL Server Compact DLL,并复制 ...
- 使用 Vscode +PlantUml 画uml图
什么是PlantUML PlantUML是一个快速创建UML图形的组件,官网上之所以称它是一个组件,主要是因为多数情况下我们都是在Eclipse.NetBenas.Intellijidea. Emac ...
- 从Uber微服务看最佳实践如何炼成?
导读:Uber成长非常迅速,工程师团队快速扩充,据说Uber有2000名工程师,8000个代码仓库,部署了1000多个微服务.微服务架构是Uber应对技术团队快速增长,功能快速上线很出色的解决方案.本 ...
- Java类的继承与方法调用的一个小问题
public class Father { protected void server(int i){ switch (i){ case 1: methodone(); break; case 2: ...
- eclipse使用lombok简化java代码
可以利用注解来代替getter.setter.toString.hashCode.构造方法等,简化代码开发. 具体用法 https://www.cnblogs.com/qnight/p/8997493 ...