Railway tickets
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 2472   Accepted: 865

Description

The railway line "Ekaterinburg-Sverdlovsk" with several stations has been built. This railway line can be represented as a line segment, railway stations being points on it. The railway line starts at the station "Ekaterinburg" and finishes at the station "Sverdlovsk", so stations are numbered starting from "Ekaterinburg" (it has number 1) and "Sverdlovsk" is the last station. 

Cost of the ticket between any two stations depends only on a distance between them. The prices for the tickets are specified in the following table.

distance between stations -X

price for the ticket

0<X<=L1

C1

L1<X<=L2

C2

L2<X<=L3

C3

Direct tickets from one station to another can be booked if and only if the distance between these station does not exceed L3. So sometimes it is necessary to book several tickets to pay for the parts of the whole way between stations.

For example, on the railway line shown at the figure above there are seven stations. The direct ticket from the second station to the sixth one can not be booked. There are several ways to pay for the travel between these stations. One of them is to book two tickets: one ticket at price C2 to travel between the second and the third stations, and other at price C3 to travel between the third and the sixth stations. Note, that though the distance between the second and the sixth stations is equal to 2*L2, the whole travel can not be paid by booking two tickets at price C2, because each ticket is valid for only one travel and each travel should start and end only at stations.

Your task is to write a program, that will find the minimal cost of the travel between two given stations.

Input

The first line of the input file contains 6 integers L1, L2, L3, C1, C2, C3 (1 <= L1 < L2 < L3 <= 10^9, 1 <= C1 < C2 < C3 <= 10^9) in the specified order with one space between. The second line contains the amount of stations N (2 <= N <= 10000). The third line contains two different integers separated by space. They represent serial numbers of stations, the travel between which must be paid. Next N-1 lines contain distances from the first station ("Ekaterinburg") on the railway line to others. These distances are given as different positive integers and are arranged in the ascending order. The distance from "Ekaterinburg" to "Sverdlovsk" does not exceed 10^9. The distance between any neighboring stations does not exceed L3. The minimal travel cost between two given stations will not exceed 10^9.

Output

Program should print to the output file the only number, which is the minimal travel cost between two given stations.

Sample Input

3 6 8 20 30 40
7
2 6
3
7
8
13
15
23

Sample Output

70
题目大意:有一个铁路线,线上有n个站,每个站之间都有一段距离,这个车站根据路程的长短出售3种票,每种票能够乘坐的距离不同,一个票只能用于从一个站到另一个站,问从站a到站b花费的最少金钱。
#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
int dist[], dp[]; int main()
{
int L1, L2, L3, C1, C2, C3;
int n, s, e;
scanf("%d%d%d%d%d%d", &L1, &L2, &L3, &C1, &C2, &C3);
scanf("%d", &n);
scanf("%d%d", &s, &e);
if (s > e)
{
s ^= e;
e ^= s;
s ^= e;
}
for (int i = s; i <= e; i++)
{
dp[i] = 0x7fffffff;
}
dist[] = ;
for (int i = ; i <= n; i++)
{
scanf("%d", &dist[i]);
}
dp[s] = ;
for (int i = s; i < e; i++)
{
for (int j = i + ; j <= e; j++)
{
if (dist[j] - dist[i] <= L1 && dp[j] > dp[i] + C1)
{
dp[j] = dp[i] + C1;
}
if (dist[j] - dist[i] <= L2 && dp[j] > dp[i] + C2)
{
dp[j] = dp[i] + C2;
}
if (dist[j] - dist[i] <= L3 && dp[j] > dp[i] + C3)
{
dp[j] = dp[i] + C3;
}
}
}
printf("%d\n", dp[e]);
return ;
}
 

POJ 2355 Railway tickets的更多相关文章

  1. poj 2828 Buy Tickets (线段树(排队插入后输出序列))

    http://poj.org/problem?id=2828 Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissio ...

  2. poj 2828 Buy Tickets 树状数组

    Buy Tickets Description Railway tickets were difficult to buy around the Lunar New Year in China, so ...

  3. poj 2828 Buy Tickets (线段树 单节点 查询位置更新)

    Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 15533   Accepted: 7759 Desc ...

  4. POJ 2828 Buy Tickets(线段树 树状数组/单点更新)

    题目链接: 传送门 Buy Tickets Time Limit: 4000MS     Memory Limit: 65536K Description Railway tickets were d ...

  5. POJ 2828 Buy Tickets

    Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get ...

  6. poj 2828 Buy Tickets【线段树单点更新】【逆序输入】

    Buy Tickets Time Limit: 4000MS   Memory Limit: 65536K Total Submissions: 16273   Accepted: 8098 Desc ...

  7. 线段树(倒序操作):POJ 2828 Buy Tickets

    Buy Tickets   Description Railway tickets were difficult to buy around the Lunar New Year in China, ...

  8. POJ 2828 Buy Tickets(排队问题,线段树应用)

    POJ 2828 Buy Tickets(排队问题,线段树应用) ACM 题目地址:POJ 2828 Buy Tickets 题意:  排队买票时候插队.  给出一些数对,分别代表某个人的想要插入的位 ...

  9. POJ - 2828 Buy Tickets (段树单点更新)

    Description Railway tickets were difficult to buy around the Lunar New Year in China, so we must get ...

随机推荐

  1. connect() to 192.168.30.71:8082 failed (99: Cannot assign requested address) while connecting to upstream, client: 114.80.182.136, server: localhost, request: "GET /home/senior HTTP/1.1", upstream: "

    connect() to 192.168.30.71:8082 failed (99: Cannot assign requested address) while connecting to ups ...

  2. 日常-acm-韩信点兵

    相传韩信才智过人,从不直接清点自己军队的人数,只要让士兵先后以三人一排,五人一排,七人一排地变换队形,而他每次只看一眼队伍的排尾就知道人数了.输入包含多组数据,每组数据包含三个非负整数a,b,c,表示 ...

  3. 2018.2.10 使用SSH连接远程滴滴云服务器Ubuntu (Windows下) 及 putty工具永久设置字体、颜色

    一开始会有人问云服务器是什么? 云服务器是一种类似VPS服务器的虚拟化技术, VPS是采用虚拟软件,VZ或VM在一台服务器上虚拟出多个类似独立服务器的部分,每个部分都可以做单独的操作系统,管理方法同服 ...

  4. CPP-网络/通信:SSL功能和原理

    面对诸多的网上交易的风险,很多网上银行.网上商城在建立系统时都注意到了安全问题.而我们也经常可以看到很多网上银行.电子商务网站在强调本系统的安全性的时候,都说采用了基于数字证书的SSL加密传输.SSL ...

  5. python之道11

    day11作业 请写出下列代码的执行结果: 例一: def func1(): print(**'in func1'**) def func2(): print(**'in func2'**) ret ...

  6. Bootstrap 标签

    本章将讲解bootstrap标签,标签可用于计数,提示和页面上其它的标记显示.使用class.laber来显示标签,如下面的实例所示 <!DOCTYPE html><html> ...

  7. iOS开发之WIFI,3G/4G两种网络同时使用技巧

    最近遇到一个比较奇葩的需求:App与硬件通过WiFi LAN通信, 同时App需要与服务器通过3G/4G WAN通信,如下图: 众所周知,手机同时打开WiFi和3G时候,会优先走WiFi.这个该如何实 ...

  8. SVN:The working copy is locked due to a previous error (一)

    使用 Cornerstone  时,碰到如题问题,SVN无法Update.Commit等操作. 解决办法:Working Copies ⟹ '右键' ⟹ Clean 即可解决! 尊重作者劳动成果,转载 ...

  9. iOS中的数据存储方式_SQLite3

    优点: 1) SQLite是一款轻型的嵌入式数据库; 2) 它占用资源非常的低,在嵌入式设备中,可能只需要几百K的内存就够了 3) 它的处理速度比Mysql.PostgreSQL这两款著名的数据库都还 ...

  10. 【离线 撤销并查集 线段树分治】bzoj1018: [SHOI2008]堵塞的交通traffic

    本题可化成更一般的问题:离线动态图询问连通性 当然可以利用它的特殊性质,采用在线线段树维护一些标记的方法 Description 有一天,由于某种穿越现象作用,你来到了传说中的小人国.小人国的布局非常 ...