洛谷 P2868 [USACO07DEC]观光奶牛Sightseeing Cows
题目描述
Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time.
Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (conveniently numbered 1.. L) and the P (2 ≤ P ≤ 5000) unidirectional cow paths that join them. Farmer John will drive the cows to a starting landmark of their choice, from which they will walk along the cow paths to a series of other landmarks, ending back at their starting landmark where Farmer John will pick them up and take them back to the farm. Because space in the city is at a premium, the cow paths are very narrow and so travel along each cow path is only allowed in one fixed direction.
While the cows may spend as much time as they like in the city, they do tend to get bored easily. Visiting each new landmark is fun, but walking between them takes time. The cows know the exact fun values Fi (1 ≤ Fi ≤ 1000) for each landmark i.
The cows also know about the cowpaths. Cowpath i connects landmark L1i to L2i (in the direction L1i -> L2i ) and requires time Ti (1 ≤ Ti ≤ 1000) to traverse.
In order to have the best possible day off, the cows want to maximize the average fun value per unit time of their trip. Of course, the landmarks are only fun the first time they are visited; the cows may pass through the landmark more than once, but they do not perceive its fun value again. Furthermore, Farmer John is making the cows visit at least two landmarks, so that they get some exercise during their day off.
Help the cows find the maximum fun value per unit time that they can achieve.
作为对奶牛们辛勤工作的回报,Farmer John决定带她们去附近的大城市玩一天。旅行的前夜,奶牛们在兴奋地讨论如何最好地享受这难得的闲暇。 很幸运地,奶牛们找到了一张详细的城市地图,上面标注了城市中所有L(2 <= L <= 1000)座标志性建筑物(建筑物按1..L顺次编号),以及连接这些建筑物的P(2 <= P <= 5000)条道路。按照计划,那天早上Farmer John会开车将奶牛们送到某个她们指定的建筑物旁边,等奶牛们完成她们的整个旅行并回到出发点后,将她们接回农场。由于大城市中总是寸土寸金,所有的道路都很窄,政府不得不把它们都设定为通行方向固定的单行道。 尽管参观那些标志性建筑物的确很有意思,但如果你认为奶牛们同样享受穿行于大城市的车流中的话,你就大错特错了。与参观景点相反,奶牛们把走路定义为无趣且令她们厌烦的活动。对于编号为i的标志性建筑物,奶牛们清楚地知道参观它能给自己带来的乐趣值F_i (1 <= F_i <= 1000)。相对于奶牛们在走路上花的时间,她们参观建筑物的耗时可以忽略不计。 奶牛们同样仔细地研究过城市中的道路。她们知道第i条道路两端的建筑物 L1_i和L2_i(道路方向为L1_i -> L2_i),以及她们从道路的一头走到另一头所需要的时间T_i(1 <= T_i <= 1000)。 为了最好地享受她们的休息日,奶牛们希望她们在一整天中平均每单位时间内获得的乐趣值最大。当然咯,奶牛们不会愿意把同一个建筑物参观两遍,也就是说,虽然她们可以两次经过同一个建筑物,但她们的乐趣值只会增加一次。顺便说一句,为了让奶牛们得到一些锻炼,Farmer John要求奶牛们参观至少2个建筑物。 请你写个程序,帮奶牛们计算一下她们能得到的最大平均乐趣值。
输入输出格式
输入格式:
Line 1: Two space-separated integers: L and P
Lines 2..L+1: Line i+1 contains a single one integer: Fi
- Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L1i , L2i , and Ti
输出格式:
- Line 1: A single number given to two decimal places (do not perform explicit rounding), the maximum possible average fun per unit time, or 0 if the cows cannot plan any trip at all in accordance with the above rules.
输入输出样例
5 7
30
10
10
5
10
1 2 3
2 3 2
3 4 5
3 5 2
4 5 5
5 1 3
5 2 2
6.00
#include <cstdio>
#include <queue>
#include <cmath>
#define N 5005
#define eps 1e-8
int n,m,f[N];
bool vis[N];
int cnt,head[N<<],nextt[N<<],to[N<<],val[N<<];
double nval[N<<],dis[N];
inline void ins(int u,int v,int l)
{
nextt[++cnt]=head[u];
to[cnt]=v;
val[cnt]=l;
head[u]=cnt;
}
void rebuild(double mid)
{
for(int i=;i<=cnt;++i)
nval[i]=(double)val[i]*mid-f[to[i]];
}
bool spfa(int x)
{
vis[x]=;
for(int i=head[x];i;i=nextt[i])
{
int v=to[i];
if(dis[v]>dis[x]+nval[i])
{
dis[v]=dis[x]+nval[i];
if(vis[v]||spfa(v))
{
vis[v]=;
return ;
}
}
}
vis[x]=;
return ;
}
bool check()
{
for(int i=;i<=n;++i) vis[i]=,dis[i]=1e8;
for(int i=;i<=n;++i)
if(spfa(i)) return ;
return ;
}
int Main()
{
freopen("sugar.in","r",stdin);freopen("sugra.out","w",stdout);
scanf("%d%d",&n,&m);
for(int i=;i<=n;++i) scanf("%d",&f[i]);
for(int x,y,z;m--;)
{
scanf("%d%d%d",&x,&y,&z);
ins(x,y,z);
}
double l=,r=,ans;
for(double mid;fabs(r-l)>=eps;)
{
mid=(l+r)/;
rebuild(mid);
if(check()) l=mid+eps,ans=mid;
else r=mid-eps;
}
printf("%.2lf\n",ans);
return ;
}
int sb=Main();
int main(int argc,char *argv[]){;}
洛谷 P2868 [USACO07DEC]观光奶牛Sightseeing Cows的更多相关文章
- 洛谷P2868 [USACO07DEC]观光奶牛Sightseeing Cows
P2868 [USACO07DEC]观光奶牛Sightseeing Cows 题目描述 Farmer John has decided to reward his cows for their har ...
- 洛谷P2868 [USACO07DEC]观光奶牛 Sightseeing Cows
题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the ...
- 洛谷P2868 [USACO07DEC]观光奶牛Sightseeing Cows(01分数规划)
题意 题目链接 Sol 复习一下01分数规划 设\(a_i\)为点权,\(b_i\)为边权,我们要最大化\(\sum \frac{a_i}{b_i}\).可以二分一个答案\(k\),我们需要检查\(\ ...
- 洛谷 P2868 [USACO07DEC]观光奶牛Sightseeing Cows 题解
题面 这道题是一道标准的01分数规划: 但是有一些细节可以优化: 不难想到要二分一个mid然后判定图上是否存在一个环S,该环是否满足∑i=1t(Fun[vi]−mid∗Tim[ei])>0 但是 ...
- POJ3621或洛谷2868 [USACO07DEC]观光奶牛Sightseeing Cows
一道\(0/1\)分数规划+负环 POJ原题链接 洛谷原题链接 显然是\(0/1\)分数规划问题. 二分答案,设二分值为\(mid\). 然后对二分进行判断,我们建立新图,没有点权,设当前有向边为\( ...
- 洛谷 2868 [USACO07DEC]观光奶牛Sightseeing Cows
题目戳这里 一句话题意 L个点,P条有向边,求图中最大比率环(权值(Fun)与长度(Tim)的比率最大的环). Solution 巨说这是0/1分数规划. 话说 0/1分数规划 是真的难,但貌似有一些 ...
- P2868 [USACO07DEC]观光奶牛Sightseeing Cows
P2868 [USACO07DEC]观光奶牛Sightseeing Cows [](https://www.cnblogs.com/images/cnblogs_com/Tony-Double-Sky ...
- [USACO07DEC]观光奶牛Sightseeing Cows 二分答案+判断负环
题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the ...
- Luogu 2868 [USACO07DEC]观光奶牛Sightseeing Cows
01分数规划复习. 这东西有一个名字叫做最优比率环. 首先这个答案具有单调性,我们考虑如何检验. 设$\frac{\sum_{i = 1}^{n}F_i}{\sum_{i = 1}^{n}T_i} = ...
随机推荐
- java POP3
package com.skyzoo.Jutil; import java.io.BufferedReader; import java.io.IOException; import java.io. ...
- jquery、javascript实现(get、post两种方式)跨域解决方法
一.实现get方式跨域请求数据 浏览器端 <script> $(document).ready(function(){ $.ajax({ url: "http://www.xxx ...
- 在UI程序设计中使用BackgroundWorker进行多线程异步处
WinForm的应用程序中如果执行了一个的非常冗长的处理操作,(比如文件检索,大运算量计算),在执行时就会锁定用户界面,虽然主活动窗口还在运行,但用户无法与程序交互,无法移动窗体或改变窗体大小,导致程 ...
- linux下 C编程改变输出字体颜色
格式: echo "\033[字背景颜色;字体颜色m字符串\033[0m"例如:echo "\033[41;36m something here \033[0m" ...
- SEO优化之——rel=canonical(整合重复页)
用法一 如图所示 有三个页面,三个页面的访问路径都不一样,但是现实的内容page2和page3显示的内容和page1一样,这时候在三个page中分别都加入了<link res="can ...
- spring boot 启动报:Composite-id class must implement Serializable: xxx错误
问题 在对spring boot项目进行修改后,重新启动时报如下错误: Composite-id class must implement Serializable: xxx xxx为具体的类 原因 ...
- java基础第五篇封装与面向对象
a.方法: public static void main(String[] args) { } 一般定义标准: 形参:一般把 不确定的量或者变化的量定义在形参位置//圆的的半径,长方形的长和宽,传递 ...
- Git 深度学习填坑之旅三(分支branch、远程操作)
0x01 分支branch依旧借用大表哥(@表元素)的图 很多时候,我们需要建立另一条分支来进行项目的独立开发,当完成后再跟主流回合进行合并这个时候就要启用分支branch功能 git branch ...
- maven scope 以及依赖传递
https://www.cnblogs.com/mxm2005/p/4947905.html
- Mysql 的 增删改查
mysql的增删改查 1:新建数据库 create database 数据库名 [其他选项]; 2:新建数据表 create table students ( id int unsigned not ...