洛谷 P3063 [USACO12DEC]牛奶的路由Milk Routing
题目背景
征求翻译。如果你能提供翻译或者题意简述,请直接发讨论,感谢你的贡献。
题目描述
Farmer John's farm has an outdated network of M pipes (1 <= M <= 500) for pumping milk from the barn to his milk storage tank. He wants to remove and update most of these over the next year, but he wants to leave exactly one path worth of pipes intact, so that he can still pump milk from the barn to the storage tank.
The pipe network is described by N junction points (1 <= N <= 500), each of which can serve as the endpoint of a set of pipes. Junction point 1 is the barn, and junction point N is the storage tank. Each of the M bi-directional pipes runs between a pair of junction points, and has an associated latency (the amount of time it takes milk to reach one end of the pipe from the other) and capacity (the amount of milk per unit time that can be pumped through the pipe in steady state). Multiple pipes can connect between the same pair of junction points.
For a path of pipes connecting from the barn to the tank, the latency of the path is the sum of the latencies of the pipes along the path, and the capacity of the path is the minimum of the capacities of the pipes along the path (since this is the "bottleneck" constraining the overall rate at which milk can be pumped through the path). If FJ wants to send a total of X units of milk through a path of pipes with latency L and capacity C, the time this takes is therefore L + X/C.
Given the structure of FJ's pipe network, please help him select a single path from the barn to the storage tank that will allow him to pump X units of milk in a minimum amount of total time.
农民约翰的农场有一套老旧的管网,管网由M条管道(1<=M<=500)构成,用于将牛奶从谷仓运到储奶罐。 他想在明年移除和更新大部分管道,但他想原封不动地保留一条完整的路径,这样他仍然可以把牛奶从谷仓输送到储罐。
管网由N个节点(1<=N<=500)组成,每个点都可以作为一组管道的端点。结点1是谷仓,结点N是储罐。M条双向管道中的每一条都连接一对节点,并且都有一个延迟值(牛奶达到管的另一端的用时)和容量值(单位时间内可以稳定通过管道的牛奶量)。多条管道可以连接同一对节点。
对于一条连接谷仓与储罐的路径,路径的延迟等于沿途所有管道的延迟之和,路径的容量等于沿途管道最小的容量(因为这是制约牛奶运送的“瓶颈”)。如果约翰通过一条延迟为L、容量为C的管道运送X个单位的牛奶,需要的时间为L+X/C。
给出约翰的管网结构,请帮助他选择一条路径,使得他从谷仓到储罐运送X个单位牛奶的总时间最少。
输入输出格式
输入格式:
* Line 1: Three space-separated integers: N M X (1 <= X <= 1,000,000).
* Lines 2..1+M: Each line describes a pipe using 4 integers: I J L C. I and J (1 <= I,J <= N) are the junction points at both ends of the pipe. L and C (1 <= L,C <= 1,000,000) give the latency and capacity of the pipe.
第1行:三个空格分隔的整数:N M X(1<=X<=1000000)。
第2行到第M+1行:每一行描述一条管道,有4个整数:I J L C。I和J(1<=I,J<=N)是这条管道连接的两个点。L和C(1<=L,C<=1000000)是这条管道的延迟和容量。
输出格式:
* Line 1: The minimum amount of time it will take FJ to send milk along a single path, rounded down to the nearest integer.
第1行:约翰沿着一条路径送牛奶花费的最少的时间,向下取整到最近的整数。
输入输出样例
说明
FJ wants to send 15 units of milk through his pipe network. Pipe #1 connects junction point 1 (the barn) to junction point 2, and has a latency of 10 and a capacity of 3. Pipes #2 and #3 are similarly defined.
The path 1->3 takes 14 + 15/1 = 29 units of time. The path 1->2->3 takes 20 + 15/2 = 27.5 units of time, and is therefore optimal.
约翰想要通过管网运送15个单位的牛奶。管道1连接节点1(谷仓)和节点2,延迟为10,容量为3。管道2和管道3也以相似的方式来定义。
路径1->3花费14+15/1=29个单位的时间。路径1->2->3花费20+15/2=27.5个单位的时间,用时最少。
思路:最短路跑一下。
#include<queue>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 5000
using namespace std;
int n,m,x,tot;
queue<int>que;
int vis[MAXN],dis[MAXN],num[MAXN];
int to[MAXN],cap[MAXN],sum[MAXN],net[MAXN],head[MAXN];
void add(int u,int v,int w,int ww){
to[++tot]=v;cap[tot]=w;sum[tot]=ww;net[tot]=head[u];head[u]=tot;
to[++tot]=u;cap[tot]=w;sum[tot]=ww;net[tot]=head[v];head[v]=tot;
}
int spfa(int s){
memset(vis,,sizeof(vis));
memset(dis,0x7f,sizeof(dis));
while(!que.empty()) que.pop();
que.push();vis[]=;dis[]=;
while(!que.empty()){
int now=que.front();
que.pop();vis[now]=;
for(int i=head[now];i;i=net[i])
if(dis[to[i]]>dis[now]+cap[i]&&sum[i]>=s){
dis[to[i]]=dis[now]+cap[i];
if(!vis[to[i]]){
vis[to[i]]=;
que.push(to[i]);
}
}
}
}
int main(){
scanf("%d%d%d",&n,&m,&x);
for(int i=;i<=m;i++){
int u,v,L,C;
scanf("%d%d%d%d",&u,&v,&L,&C);
add(u,v,L,C);num[i]=C;
}
sort(num+,num++m);
int minn=0x7f7f7f7f;
for(int i=;i<=m;i++){
spfa(num[i]);
minn=min(minn,dis[n]+x/num[i]);
}
cout<<minn;
}
洛谷 P3063 [USACO12DEC]牛奶的路由Milk Routing的更多相关文章
- 洛谷P3063 [USACO12DEC]牛奶的路由Milk Routing
链接 其实在博客园里写题解都挺应付的都是在洛谷写了之后 挑一部分粘过来 在洛谷写的也都是废话,是为了凑篇幅 主要就是代码 大体思路就一提 这题贪心不行废话 跑m遍SPFA更新最小值 注意数组记得清空 ...
- 【luogu P3063 [USACO12DEC]牛奶的路由Milk Routing】 题解
题目链接:https://www.luogu.org/problemnew/show/P3063#sub 我很好奇这道题为什么没被收入SPFA好题 #include <cstdio> #i ...
- [洛谷P2852] [USACO06DEC]牛奶模式Milk Patterns
洛谷题目链接:[USACO06DEC]牛奶模式Milk Patterns 题目描述 Farmer John has noticed that the quality of milk given by ...
- 洛谷 P3063 【[USACO12DEC]Milk Routing S】
这道题可以暴力哒~ 我们枚举每一个出现过的容量,然后跑一次最短路,求延迟,在跑最短路的时候,如果遇到的某一个点,比我们当前枚举的那个点小,那么就直接不走这一个点,然后枚举完后,就能得到最大值了. 代码 ...
- 洛谷P3093 [USACO13DEC]牛奶调度Milk Scheduling
题目描述 Farmer John has N cows that need to be milked (1 <= N <= 10,000), each of which takes onl ...
- 洛谷P1208——P1208 [USACO1.3]Mixing Milk(贪心)
题目描述 由于乳制品产业利润很低,所以降低原材料(牛奶)价格就变得十分重要.帮助Marry乳业找到最优的牛奶采购方案. Marry乳业从一些奶农手中采购牛奶,并且每一位奶农为乳制品加工企业提供的价格是 ...
- 洛谷 P1208混合牛奶【贪心】
题目描述 由于乳制品产业利润很低,所以降低原材料(牛奶)价格就变得十分重要.帮助Marry乳业找到最优的牛奶采购方案. Marry乳业从一些奶农手中采购牛奶,并且每一位奶农为乳制品加工企业提供的价格是 ...
- 洛谷P3065 [USACO12DEC]第一!First!(Trie树+拓扑排序)
P3065 [USACO12DEC]第一!First! 题目链接:https://www.luogu.org/problemnew/show/P3065 题目描述 Bessie一直在研究字符串.她发现 ...
- 洛谷P3066 [USACO12DEC]逃跑的BarnRunning Away From…
题面链接 一句话题意:给出以1号点为根的一棵有根树,问每个点的子树中与它距离小于等于l的点有多少个. 我:似乎并不好做啊...看了题解后大雾... sol:考虑树上差分,对于一个点,在他那个位置++, ...
随机推荐
- ACM_01背包
背包1 Time Limit: 2000/1000ms (Java/Others) Problem Description: 有n个重量和价值分别为Wi,Vi的物品,现从这些物品中挑选出总量不超过 W ...
- pyinstaller打包报错:AttributeError: 'str' object has no attribute 'items'
导致原因和python多数奇奇怪怪的问题一样,依赖包的版本问题. 解决办法: 对setuptools这个包进行升级,链接在这里 https://pypi.org/project/setuptools/ ...
- swiper3初始化/swiper-init/用data实例化swiper/data-swiper
Framework7直接用data属性实例化swiper用起来很爽,刚好最近又用到swiper插件,自己写一个 HTML <div class="swiper-container sw ...
- java 物理分页和逻辑分页
A.逻辑分页利用游标分页,好处是所有数据库都统一,坏处就是效率低.1.逻辑分页的第一种方式,利用ResultSet的滚动分页.这种分页方式依靠的是对结果集的算法来分页,因此通常被称为“逻辑分页”.步骤 ...
- 自学php【一】 任务:图片上传即时可见
工作已经快2周了,头儿给派了个任务做个企业站!这几天正在紧锣密鼓的作战中!等忙完了这个活!写下自己的学习心得体会!与看到文章的您一起分享! 在这里记录每次遇到的难题,如何解决的! 今天要做的功能就是实 ...
- sublime text3 =个人插件
1.sublime text3汉化插件安装. ctrl+shift+p → Package Control:Install Package → ChineseLocalization preferen ...
- Python+selenium学习(二) selenium 定位不到元素
转载:https://www.cnblogs.com/tarrying/p/9681991.html tarrying selenium的三种等待时间 //隐式等待(20秒以内没哥一段时间就会去找元素 ...
- ERwin 正向工程
1.物理模型带字段备注 COMMENT ON 将模型切换至 physical 模式选择 Model ---> Domain Dictionary , 在 tab 标签中,切换至comment 然 ...
- JAVA程序员面试笔试宝典4
1.HTTP中GET与POST方法有什么区别? GET方法上传数据时,数据添加在URL后面.同时,数据大小有限制,通常在1024Byte左右.POST方法传递数据是通过HTTP请求的附件进行的,传递的 ...
- JavaScript--小白入门篇2
一.布尔值和关系运算符.逻辑运算符 1.1 布尔值 我们上篇文章说了,学习了两种变量的类型数值型.字符串型. 实际上,还有很多变量的类型.我们今天再学习一种,叫做“布尔类型”. 数值型里面的值 ...