解题报告

题意:

有一个旅游团如今去出游玩,如今有n个城市,m条路。因为每一条路上面规定了最多可以通过的人数,如今想问这个旅游团人数已知的情况下最少须要运送几趟

思路:

求出发点到终点全部路其中最小值最大的那一条路。

求发可能有多种。最短路的松弛方式改掉是一种。最小生成树的解法也是一种(ps。prime和dijs就是这样子类似的)

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define inf 0x3f3f3f3f
using namespace std;
int n,m,q,mmap[110][110];
void floyd() {
for(int k=0; k<n; k++)
for(int i=0; i<n; i++)
for(int j=0; j<n; j++)
mmap[i][j]=max(mmap[i][j],min(mmap[i][k],mmap[k][j]));
}
int main() {
int i,j,u,v,w,k=1;
while(~scanf("%d%d",&n,&m)) {
if(!n&&!m)break;
for(i=0; i<n; i++) {
for(j=0; j<n; j++)
mmap[i][j]=0;
}
for(i=0; i<m; i++) {
scanf("%d%d%d",&u,&v,&w);
mmap[u-1][v-1]=mmap[v-1][u-1]=w;
}
floyd();
scanf("%d%d%d",&u,&v,&w);
int ans=0,num=mmap[u-1][v-1]-1;
ans=ceil((double)w/num);
printf("Scenario #%d\n",k++);
printf("Minimum Number of Trips = %d\n\n",ans);
}
return 0;
}

The Tourist Guide

Input: standard input

Output: standard output

Mr. G. works as a tourist guide. His current assignment is to take some tourists from one city to another. Some two-way roads connect the cities. For each pair of neighboring cities there is a bus service that runs only between those two cities and uses
the road that directly connects them. Each bus service has a limit on the maximum number of passengers it can carry. Mr. G. has a map showing the cities and the roads connecting them. He also has the information regarding each bus service. He understands that
it may not always be possible for him to take all the tourists to the destination city in a single trip. For example, consider the following road map of 7 cities. The edges connecting the cities represent the roads and the number written on each edge indicates
the passenger limit of the bus service that runs on that road.

Now, if he wants to take 99 tourists from city 1 to city 7, he will require at least 5 trips, since he has to ride the bus with each group, and the route he should take is : 1 - 2 - 4 - 7.

But, Mr. G. finds it difficult to find the best route all by himself so that he may be able to take all the tourists to the destination city in minimum number of trips. So, he seeks your help.

 

Input

The input will contain one or more test cases. The first line of each test case will contain two integers: N (N<= 100) and R representing respectively the number of cities and the number of road segments. Then R lines will follow
each containing three integers: C1C2 andPC1 and C2 are
the city numbers and P (P> 1) is the limit on the maximum number of passengers to be carried by the bus service between the two cities. City numbers are positive integers ranging from 1 to N. The (R + 1)-th line will contain three
integers: SD andT representing respectively the starting city, the destination city and the number of tourists to be guided.

The input will end with two zeroes for N and R.

 

Output

For each test case in the input first output the scenario number. Then output the minimum number of trips required for this case on a separate line. Print a blank line after the output of each test case.

 

Sample Input

7 10

1 2 30

1 3 15

1 4 10

2 4 25

2 5 60

3 4 40

3 6 20

4 7 35

5 7 20

6 7 30

1 7 99

0 0

Sample Output

Scenario #1

Minimum Number of Trips = 5

UVa10099_The Tourist Guide(最短路/floyd)(小白书图论专题)的更多相关文章

  1. UVa10048_Audiophobia(最短路/floyd)(小白书图论专题)

    解题报告 题意: 求全部路中最大分贝最小的路. 思路: 类似floyd算法的思想.u->v能够有另外一点k.通过u->k->v来走,拿u->k和k->v的最大值和u-&g ...

  2. UVa567_Risk(最短路)(小白书图论专题)

    解题报告 option=com_onlinejudge&Itemid=8&category=7&page=show_problem&problem=508"& ...

  3. UVa753/POJ1087_A Plug for UNIX(网络流最大流)(小白书图论专题)

    解题报告 题意: n个插头m个设备k种转换器.求有多少设备无法插入. 思路: 定义源点和汇点,源点和设备相连,容量为1. 汇点和插头相连,容量也为1. 插头和设备相连,容量也为1. 可转换插头相连,容 ...

  4. UVa563_Crimewave(网络流/最大流)(小白书图论专题)

    解题报告 思路: 要求抢劫银行的伙伴(想了N多名词来形容,强盗,贼匪,小偷,sad.都认为不合适)不在同一个路口相碰面,能够把点拆成两个点,一个入点.一个出点. 再设计源点s连向银行位置.再矩阵外围套 ...

  5. UVa10397_Connect the Campus(最小生成树)(小白书图论专题)

    解题报告 题目传送门 题意: 使得学校网络互通的最小花费,一些楼的线路已经有了. 思路: 存在的线路当然全都利用那样花费肯定最小,把存在的线路当成花费0,求最小生成树 #include <ios ...

  6. UVa409_Excuses, Excuses!(小白书字符串专题)

    解题报告 题意: 找包括单词最多的串.有多个按顺序输出 思路: 字典树爆. #include <cstdio> #include <cstring> #include < ...

  7. floyd类型题UVa-10099-The Tourist Guide +Frogger POJ - 2253

    The Tourist Guide Mr. G. works as a tourist guide. His current assignment is to take some tourists f ...

  8. [uva] 10099 - The Tourist Guide

    10099 - The Tourist Guide 题目页:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemi ...

  9. ACM/ICPC 之 最短路-Floyd+SPFA(BFS)+DP(ZOJ1232)

    这是一道非常好的题目,融合了很多知识点. ZOJ1232-Adventrue of Super Mario 这一题折磨我挺长时间的,不过最后做出来非常开心啊,哇咔咔咔 题意就不累述了,注释有写,难点在 ...

随机推荐

  1. NSNotificationCenter监听TextField文字变化

    注册 1: NSNotificationCenter.defaultCenter().addObserver(self, selector: "textDidChange", na ...

  2. Swift数独游戏优化——C++与OC混编、plist自动生成

    一.为什么要C++与OC混编? 在我之前的数独游戏中涉及到的数独游戏生成算法是参考的网上其他人的算法,是利用C++来实现的.   但是在我的例子中我发现这样存在一定的局限性: 1.我是利用Termin ...

  3. linux下安装python2.7.5和MYSQLdb

    由于开发的python web 扫描器需要在python2.7.5以及需要MYSQLdb这个库的支持,在此做一个记录,避免更换到新环境时的学习成本. 一.安装MYSQL 1.yum install m ...

  4. 关于css解决俩边等高的问题(等高布局)

    等高布局 前段时间公司需哦一个后台管理系统,左侧是导航栏,右侧是content区域.然厚刚开始用的是js 去控制的,但是当页面的椰蓉过长的时候,有与js单线程,加载比较慢,就会有那么一个过程,查找了很 ...

  5. flask_admin model官方文档学习

    文档地址:http://www.minzhulou.com/docs/flask-admin/api/mod_model.html model在flask_admin算是比较重要的部分,根据文档稍微的 ...

  6. python 页面信息抓取

    1. 特点 在python 解析html这篇文章中已经做了初步的介绍,接下来再坐进一步的说明.python抓取页面信息有下面两个特点: 依赖于HTML的架构. 微小的变化可能会导致抓取失败,这取决于你 ...

  7. android studio 警告 synchronization on non-final field

    测试代码如下: public class SyncNonFinalField { private Object object = new Object(); public void start() { ...

  8. POJ1274:The Perfect Stall(二分图最大匹配 匈牙利算法)

    The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17895   Accepted: 814 ...

  9. iOS开发之实现半透明蒙层背景效果[用于下拉菜单页和分享页]

    郝萌主倾心贡献.尊重作者的劳动成果,请勿转载. 假设文章对您有所帮助.欢迎给作者捐赠.支持郝萌主,捐赠数额任意.重在心意^_^ 我要捐赠: 点击捐赠 Cocos2d-X源代码下载:点我传送 游戏官方下 ...

  10. Java 创建用户异常类、将异常一直向上抛、 throw和throws的区别

    如果java提供的系统异常类型不能满足程序设计的需求,那么可以设计自己的异常类型. 从java异常类的结构层次可以看出,java类型的公共父类为Throwable.在程序运行中可能出现俩种问题:一种是 ...