UVA 507 - Jill Rides Again 动态规划
| Jill Rides Again |
Jill likes to ride her bicycle, but since the pretty city of Greenhills where she lives has grown, Jill often uses the excellent public bus system for part of her journey. She has a folding bicycle which she carries with her when she uses the bus for the first part of her trip. When the bus reaches some pleasant part of the city, Jill gets off and rides her bicycle. She follows the bus route until she reaches her destination or she comes to a part of the city she does not like. In the latter event she will board the bus to finish her trip.
Through years of experience, Jill has rated each road on an integer scale of ``niceness.'' Positive niceness values indicate roads Jill likes; negative values are used for roads she does not like. There are not zero values. Jill plans where to leave the bus and start bicycling, as well as where to stop bicycling and re-join the bus, so that the sum of niceness values of the roads she bicycles on is maximized. This means that she will sometimes cycle along a road she does not like, provided that it joins up two other parts of her journey involving roads she likes enough to compensate. It may be that no part of the route is suitable for cycling so that Jill takes the bus for its entire route. Conversely, it may be that the whole route is so nice Jill will not use the bus at all.
Since there are many different bus routes, each with several stops at which Jill could leave or enter the bus, she feels that a computer program could help her identify the best part to cycle for each bus route.
Input
The input file contains information on several bus routes. The first line of the file is a single integer
b
representing the number of route descriptions in the file. The identifier for each route (
r
) is the sequence number within the data file,

. Each route description begins with the number of stops on the route: an integer
s
,

on a line by itself. The number of stops is followed by
s
- 1 lines, each line
i
(

) is an integer
n
i
representing Jill's assessment of the niceness of the road between the two stops
i
and
i
+1.
Output
For each route
r
in the input file, your program should identify the beginning bus stop
i
and the ending bus stop
j
that identify the segment of the route which yields the maximal sum of niceness, m= n
i
+n
i+1
+...+n
j-1
. If more than one segment is maximally nice, choose the one with the longest cycle ride (largest
j
-
i
). To break ties in longest maximal segments, choose the segment that begins with the earliest stop (lowest
i
). For each route
r
in the input file, print a line in the form:
The nicest part of route r is between stops i and j
However, if the maximal sum is not positive, your program should print:
Route r has no nice parts
Sample Input
3
3
-1
6
10
4
-5
4
-3
4
4
-4
4
-5
4
-2
-3
-4
Sample Output
The nicest part of route 1 is between stops 2 and 3
The nicest part of route 2 is between stops 3 and 9
Route 3 has no nice parts
题意: 一段公交车路,各个车站为1,2,3...s, 各个车站之间的景色值是不同的, 例如车站1到车站2的景色值是5, 车站3到车站4的景色值是-3. 求一段连续的车站的景色值之和最大是多少。
#include<stdio.h>
#include<string.h>
int l[20005];
int main()
{
int b,n,i,max,sum,st,ed,cas=1,bg;
scanf("%d",&b);
while(b--)
{
scanf("%d",&n);
for(i=1;i<n;i++)
scanf("%d",&l[i]);
max=sum=l[1];
st=bg=ed=1; /*开始时起点和终点都在第1站*/
for(i=2;i<n;i++)
{
if(sum<0)
{
sum=0;
st=i; /*更新起点*/
}
sum+=l[i];
if(sum>max||sum==max&&i-st>ed-bg)
{
max=sum;
bg=st; /*bg总是保存新的起点*/
ed=i;
}
}
if(max<=0)
printf("Route %d has no nice parts\n",cas++);
else
printf("The nicest part of route %d is between stops %d and %d\n",cas++,bg,ed+1);
}
return 0;
}
UVA 507 - Jill Rides Again 动态规划的更多相关文章
- UVa 507 - Jill Rides Again
题目大意:最大和子序列问题.由于具有最大和的子序列具有一下性质:第一项不为负数,并且从第一项开始累加,中间不会有和出现负数,因为一旦有负数我们可以抛弃前边的部分以得到更大的子序列和,这将会产生矛盾. ...
- CJOJ 1070 【Uva】嵌套矩形(动态规划 图论)
CJOJ 1070 [Uva]嵌套矩形(动态规划 图论) Description 有 n 个矩形,每个矩形可以用两个整数 a, b 描述,表示它的长和宽.矩形 X(a, b) 可以嵌套在矩形 Y(c, ...
- CJOJ 1071 【Uva】硬币问题(动态规划)
CJOJ 1071 [Uva]硬币问题(动态规划) Description 有n种硬币,面值分别为v1, v2, ..., vn,每种都有无限多.给定非负整数S,可以选用多少个硬币,使得面值之和恰好为 ...
- uva507 - Jill Rides Again(最长连续和)
option=com_onlinejudge&Itemid=8&page=show_problem&problem=448">题目:uva507 - Jill ...
- UVa 104 - Arbitrage(Floyd动态规划)
题目来源:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=3&pa ...
- UVa 10891 - Game of Sum 动态规划,博弈 难度: 0
题目 https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&a ...
- UVa 10891 Game of Sum - 动态规划
因为数的总和一定,所以用一个人得分越高,那么另一个人的得分越低. 用$dp[i][j]$表示从$[i, j]$开始游戏,先手能够取得的最高分. 转移通过枚举取的数的个数$k$来转移.因为你希望先手得分 ...
- UVa 10635 Prince and Princess - 动态规划
讲一下题目大意,就是有两个长度为p + 1和q + 1的序列,求它们的LCS. 如果用O(pq)的算法对于这道题来说还是太慢了.所以要另外想一些方法.注意到序列中的所有元素都不相同,所以两个序列中数对 ...
- uva 116 Unidirectional TSP(动态规划,多段图上的最短路)
这道题目并不是很难理解,题目大意就是求从第一列到最后一列的一个字典序最小的最短路,要求不仅输出最短路长度,还要输出字典序最小的路径. 这道题可以利用动态规划求解.状态定义为: cost[i][j] = ...
随机推荐
- error LNK2019
error LNK2019: 无法解析的外部符号 "public: virtual __thiscall Fruit::~Fruit(void)" (??1Fruit@@UAE@X ...
- 为什么学微信小程序开发
微信小程序是什么? 触手可得,不用安装,不体验过自己是想不到的 ---张小龙 小程序是一种不需要下载安装即可使用的应用,它实现了应用“触手可及”的梦想,用户扫一扫或者搜 一下即可打开应用.也体现了“用 ...
- ssh登录命令(转)
转:http://blog.csdn.net/edward_qing_lee/article/details/23133331 常用格式:ssh [-l login_name] [-p port] [ ...
- 基于内容的自适应变长编码[CAVLC]
基于内容自适应的变长编码方式用于编码zigzag顺序扫描的4x4和2x2残差变换系数块. 1.编码系数个数和零序列(coeff_token): coeff_token = <TotalCoeff ...
- Echarts动态数据显示
自己慢慢摸索出来的,留着以后可能会用到 一.先引入jquery,再引入echarts.js 二.配置容器 三.配置路径和数据选项等 <script type="text/javascr ...
- 依赖注入 | Dependency Injection
原文链接: Angular Dependency Injection翻译人员: 铁锚翻译时间: 2014年02月10日说明: 译者认为,本文中所有名词性的"依赖" 都可以理解为 & ...
- 初识HTML5
1, 新增canvas标签,允许通过JS在客户端完成2D绘图 2, 新增Video/Audio标签,能取代flash实现媒体播放 3, 新增本地存储功能:localStorage/sessionSto ...
- PHP实现简单爬虫
<?php /** * 爬虫程序 -- 原型 * * 从给定的url获取html内容 * * @param string $url * @return string */ func ...
- BaseActivity的定义——作为所有Activity类的父类
public abstract class BaseActivity extends AppCompatActivity implements View.OnClickListener { prote ...
- lda模型的python实现
LDA(Latent Dirichlet Allocation)是一种文档主题生成模型,最近看了点资料,准备使用python实现一下.至于数学模型相关知识,某度一大堆,这里也给出之前参考过的一个挺详细 ...