Einbahnstrasse
Einbahnstrasse |
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
Total Submission(s): 124 Accepted Submission(s): 54 |
Problem Description
Einbahnstra e (German for a one-way street) is a street on which vehicles should only move in one direction. One reason for having one-way streets is to facilitate a smoother flow of traffic through crowded areas. This is useful in city centers, especially old cities like Cairo and Damascus. Careful planning guarantees that you can get to any location starting from any point. Nevertheless, drivers must carefully plan their route in order to avoid prolonging their trip due to one-way streets. Experienced drivers know that there are multiple paths to travel between any two locations. Not only that, there might be multiple roads between the same two locations. Knowing the shortest way between any two locations is a must! This is even more important when driving vehicles that are hard to maneuver (garbage trucks, towing trucks, etc.)
You just started a new job at a car-towing company. The company has a number of towing trucks parked at the company's garage. A tow-truck lifts the front or back wheels of a broken car in order to pull it straight back to the company's garage. You receive calls from various parts of the city about broken cars that need to be towed. The cars have to be towed in the same order as you receive the calls. Your job is to advise the tow-truck drivers regarding the shortest way in order to collect all broken cars back in to the company's garage. At the end of the day, you have to report to the management the total distance traveled by the trucks. |
Input
Your program will be tested on one or more test cases. The first line of each test case specifies three numbers (N , C , and R ) separated by one or more spaces. The city has N locations with distinct names, including the company's garage. C is the number of broken cars. R is the number of roads in the city. Note that 0 < N < 100 , 0<=C < 1000 , and R < 10000 . The second line is made of C + 1 words, the first being the location of the company's garage, and the rest being the locations of the broken cars. A location is a word made of 10 letters or less. Letter case is significant. After the second line, there will be exactly R lines, each describing a road. A road is described using one of these three formats:
A -v -> B A and B are names of two different locations, while v is a positive integer (not exceeding 1000) denoting the length of the road. The first format specifies a one-way street from location A to B , the second specifies a one-way street from B to A , while the last specifies a two-way street between them. A , ``the arrow", and B are separated by one or more spaces. The end of the test cases is specified with a line having three zeros (for N , C , and R .) The test case in the example below is the same as the one in the figure. |
Output
For each test case, print the total distance traveled using the following format:
k . V Where k is test case number (starting at 1,) is a space, and V is the result. |
Sample Input
4 2 5 |
Sample Output
1. 80 |
Source
2008 ANARC
|
Recommend
lcy
|
/*
求从总公司出发按顺序访问各个点然后回到公司的最小距离 Floyd 是固定松弛点在放缩,想错了
floy算法最简便
*/
#include<bits/stdc++.h>
#define N 105
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
int n,c,m,cur;
int mapn[N][N];
map<string,int>ma;
char city[][N],u[N],v[N],opl,opr;
int val;
/*
void floyd(){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
for(int k=1;k<=n;k++){
mapn[i][j]=min(mapn[i][j],mapn[i][k]+mapn[k][j]);
}
}
}
}
*/
/*
手写模板错了,应该固定松弛的点然后,在更新状态
*/
void floyd()
{
int i,j,k;
for(i = ; i<=n; i++)
for(j = ; j<=n; j++)
for(k = ; k<=n; k++)
if(mapn[j][i]+mapn[i][k]<mapn[j][k])
mapn[j][k] = mapn[j][i]+mapn[i][k];
} void init(){
for(int i=;i<=n;i++){
for(int j=;j<=n;j++){
mapn[i][j]=INF;
}
}
}
int main(){
//freopen("C:\\Users\\acer\\Desktop\\in.txt","r",stdin);
int Case=;
while(scanf("%d%d%d",&n,&c,&m)!=EOF&&(n+c+m)){
init();
ma.clear();
for(int i=;i<=c;i++){
scanf("%s",city[i]);
//cout<<city<<" ";
}
//cout<<endl;
int len=;//表示结点(这个地方坐标必须从1开始,因为map中没有的东西会返回的键值是0)
for(int i=;i<m;i++){
scanf("%s %c-%d-%c %s",u,&opl,&val,&opr,v);
//printf("%s %c-%d-%c %s\n",u,opl,val,opr,v);
/*
利用map的性质,没有插入的元素的第二个键值为0
*/
if(!ma[u]){//在城市中没有标记
ma[u]=len++;
//cout<<u<<endl;
}
if(!ma[v]){//在城市中没有标记
ma[v]=len++;
//cout<<v<<endl;
}
if(opl=='<'){
if(mapn[ma[v]][ma[u]]>val)
mapn[ma[v]][ma[u]]=val;
}
if(opr=='>'){
if(mapn[ma[u]][ma[v]]>val)
mapn[ma[u]][ma[v]]=val;
}
}//建图
//cout<<len<<endl;
//for(int i=0;i<n;i++){
// for(int j=0;j<n;j++){
// cout<<mapn[i][j]<<" ";
// }
// cout<<endl;
//}
floyd();
cur=;
int lost=ma[city[]];
for(int i=;i<=c;i++){
cur+=mapn[lost][ma[city[i]]]+mapn[ma[city[i]]][lost];
//cout<<mapn[lost][ma[city[i]]]<<endl;
}
printf("%d. %d\n",Case++,cur);
}
return ;
}
Einbahnstrasse的更多相关文章
- HDU2923 Einbahnstrasse (Floyd)
Einbahnstrasse Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)To ...
- HDU - 2923 - Einbahnstrasse
题目: Einbahnstrasse Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Other ...
- Einbahnstrasse HDU2923
基础2923题 处理输入很麻烦 有可能一个城市有多辆破车要拖 应该严谨一点的 考虑所有情况 #include<bits/stdc++.h> using namespace std; ] ...
- 【转】最短路&差分约束题集
转自:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★254 ...
- 【转载】图论 500题——主要为hdu/poj/zoj
转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并 ...
- 【HDOJ图论题集】【转】
=============================以下是最小生成树+并查集====================================== [HDU] How Many Table ...
- hdu图论题目分类
=============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many ...
- 转载 - 最短路&差分约束题集
出处:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★ ...
- 最短路&查分约束
[HDU] 1548 A strange lift 根蒂根基最短路(或bfs)★ 2544 最短路 根蒂根基最短路★ 3790 最短路径题目 根蒂根基最短路★ 2066 一小我的观光 根蒂根基最短路( ...
随机推荐
- mvc一对多模型表单的快速构建
功能需求描述 Q:在实际的开发中,经常会遇到一个模型中包含有多个条目的表单.如何将数据提交到后台? A: 以数组的形式提交到后台就Ok了(真的那么简单么,如果再嵌套一层呢?) A2:拆分多个模型,映射 ...
- webstorm配置scss环境
1.下载 Ruby (安装过程中记得勾选添加到环境变量,安装结束最后可能会弹出一个cmd弹框,可以忽略) 2. cmd安装sass gem install sass 3. cmd检查是否安装 sas ...
- Finally-操作返回值
Finally中操作返回会出现一个问题?直接看代码 static int M1() { ; try { result = result + ; //======引发异常的代码========== , ...
- CentOS更新源
1.首先备份/etc/yum.repos.d/CentOS-Base.repo mv /etc/yum.repos.d/CentOS-Base.repo /etc/yum.repos.d/CentOS ...
- jsonp其实很简单【ajax跨域请求】
js便签笔记(13)——jsonp其实很简单[ajax跨域请求] 前两天被问到ajax跨域如何解决,还真被问住了,光知道有个什么jsonp,迷迷糊糊的没有说上来.抱着有问题必须解决的态度,我看了许多资 ...
- Spring 学习——基于Spring WebSocket 和STOMP实现简单的聊天功能
本篇主要讲解如何使用Spring websocket 和STOMP搭建一个简单的聊天功能项目,里面使用到的技术,如websocket和STOMP等会简单介绍,不会太深,如果对相关介绍不是很了解的,请自 ...
- 【转】TCP/IP协议中TCP和UDP的区别
TCP协议与UDP协议的区别 首先咱们弄清楚,TCP协议和UCP协议与TCP/IP协议的联系,很多人犯糊涂了,一直都是说TCP/IP协议与UDP协议的区别,我觉得这是没有从本质上弄清楚网络通信! ...
- asp.net web api 2.2 基础框架(带例子)
链接:https://github.com/solenovex/asp.net-web-api-2.2-starter-template 简介 这个是我自己编写的asp.net web api 2.2 ...
- 数据分析前戏:ipython使用技巧(上)
不一定非得使用Jupyter Notebook,试试ipython命令行 安装 ipython 我只试过Windows 10环境下的. 1.安装python安装包之后,应该就有ipython了. 2. ...
- Android02-控件
在android studio中,新建一个module时布局文件中就会默认带一个TextView,里面显示着一句话:Hello World ! 布局中通常放置的是android控件,下面介绍几个an ...