UVa Problem 10187 From Dusk till Dawn （从黄昏到拂晓）

// From Dusk till Dawn （从黄昏到拂晓）// PC/UVa IDs: 110907/10187, Popularity: B, Success rate: average Level: 3// Verdict: Accepted// Submission Date: 2011-10-02// UVa Run Time: 0.048s//// 版权所有（C）2011，邱秋。metaphysis # yeah dot net//// [问题描述]// Vladimir has white skin, very long teeth and is 600 years old, but this is no// problem because Vladimir is a vampire.// Vladimir has never had any problems with being a vampire. In fact, he is a very// successful doctor who always takes the night shift and so has made many friends// among his colleagues. He has a very impressive trick which he shows at dinner// partys: He can tell tell blood group by taste.// Vladimir loves to travel, but being a vampire he has to overcome three problems.//// First, he can only travel by train because he has to take his coffin with him.// (On the up side he can always travel first class because he has invested a lot// of money in long term stocks.)// Second, he can only travel from dusk till dawn, namely from 6 pm to 6 am. During// the day he has to stay inside a train station.// Third, he has to take something to eat with him. He needs one litre of blood// per day, which he drinks at noon (12:00) inside his coffin. //// You should help Vladimir to find the shortest route between two given cities,// so that he can travel with the minimum amount of blood. (If he takes too much// with him, people will ask funny questions like "What do you do with all that// blood?")//// [输入]// The first line of the input will contain a single number telling you the number// of test cases.// Each test case specification begins with a single number telling you how many// route specifications follow.// Each route specification consists of the names of two cities, the departure// time from city one and the total travelling time. The times are in hours. Note// that Vladimir can't use routes departing earlier than 18:00 or arriving later// than 6:00.// There will be at most 100 cities and less than 1000 connections. No route takes// less than one hour and more than 24 hours. (Note that Vladimir can use only// routes with a maximum of 12 hours travel time (from dusk till dawn).) All city// names are shorter than 32 characters.// The last line contains two city names. The first is Vladimir's start city, the// second is Vladimir's destination. //// [输出]// For each test case you should output the number of the test case followed by// "Vladimir needs # litre(s) of blood." or "There is no route Vladimir can take." //// [样例输入]// 2// 3// Ulm Muenchen 17 2// Ulm Muenchen 19 12// Ulm Muenchen 5 2// Ulm Muenchen// 10// Lugoj Sibiu 12 6// Lugoj Sibiu 18 6// Lugoj Sibiu 24 5// Lugoj Medias 22 8// Lugoj Medias 18 8// Lugoj Reghin 17 4// Sibiu Reghin 19 9// Sibiu Medias 20 3// Reghin Medias 20 4// Reghin Bacau 24 6// Lugoj Bacau//// [样例输出]// Test Case 1.// There is no route Vladimir can take.// Test Case 2.// Vladimir needs 2 litre(s) of blood.//// [解题方法]// 由题意可知，Vladimir 在乘坐火车从一个城市到另一个城市时可以有多种路线，其中出发时间不在 18：00// 之后且到达时间不在 06：00 之前的都可以不予考虑，然后对这些满足要求的路线用 DAG 建模，问题即求// 图中两点的最短路线。其中需要注意的是，由于需要将携带血量最小化，故需要考虑从起始城市到终点城市的// 所有路线，因为有可能从 A 到 B 的车是 18:00 出发，20:00 到达 B，然后再从 B 到 C，有出发时间// 为 21:00，24:00 到达 C 的火车路线，这样若从 A 到 C 就不需在车站停留，从而不需喝 1 升的血，// 所以虽然总路线不是最短的，但是携带血量可以是最少的，故需要遍历可能的通路来获取最少使用血量，这个// 可以通过使用宽度优先遍历的思想来解决。#include <iostream>#include <queue>#include <map>using namespace std;#define MAXN 100#define EARLIER 18#define LATER 6#define HOURS 24#define NO_ROUTE (-1)// 当前路线的状态，其中 city 标志当前到达的城市，time 表示到达的时间，litres 表示已经使用的血量。class state{public:int city, time, litres;// 使用血量少的路线先处理。bool operator<(const state ¤t) const{return litres > current.litres;}};// 火车路线。class route{public:int departure;// 出发时间。int arrive;// 到达时间。int to;// 到达城市。};// 城市之间的火车路线。vector < route > edges[MAXN + 1];// 使用优先队列的方法来遍历所有从起始城市 from 到 终点城市 to 的路线。在此过程中，先处理使用血// 量少的路线，当发现当前路线的状态 state 所标志的城市 city 已经为目标城市 to 时，表明当前需要// 血量最少的路线即为该 state 所走的路线。int travel(int from, int to){// 建立优先队列，需要血量少的路线先处理。priority_queue < state > states;// 将当前城市为 from，到达时间为 18:00 的状态添加到优先队列中，表示已经从一个虚拟的地// 方到达了起始城市 from，到达时间小于起始城市 from 的所有可用火车路线的出发时间，当前// 已经使用血量为 0 升。states.push((state){from, EARLIER, 0});// 处理队列中的路线状态。while (!states.empty()){// 取出队首的路线状态处理。state current = states.top();states.pop();// 若当前城市已经为目标城市，则直接返回当前路线所使用的血量。if (current.city == to)return current.litres;// 遍历当前到达城市至其他城市的火车路线，根据情况决定是否需要取用 1 升的血。for (int r = 0; r < edges[current.city].size(); r++){int used = current.litres;// 若到达时间晚于该火车路线出发时间，则需在此车站停留一个中午，取血 1 升。if (current.time > edges[current.city][r].departure)used++;states.push((state){edges[current.city][r].to,edges[current.city][r].arrive, used});}}// 起始城市与目标城市无连通路线。return NO_ROUTE;}int main(int ac, char *av[]){int test, routes, cases = 1;int from, to;string start, destination;int departure, traveling;cin >> test;while (test--){map < string, int > cities;for (int i = 0; i < MAXN + 1; i++)edges[i].clear();cin >> routes;for (int i = 1; i <= routes; i++){// 读入起始和终点城市、出发时间、旅行时间，并将其插入到 map 中。cin >> start >> destination >> departure >> traveling;if (cities.find(start) == cities.end()){from = cities.size();cities[start] = from;}elsefrom = cities[start];if (cities.find(destination) == cities.end()){to = cities.size();cities[destination] = to;}elseto = cities[destination];// 满足条件的路线则添加到有向图中。若出发时间为凌晨，则加上 24 小时// 以统一时间起点。departure += (departure <= LATER ? HOURS : 0);if (departure >= EARLIER&& (departure + traveling) <= (HOURS + LATER))edges[from].push_back((route){departure,(departure + traveling), to});}// 出发和到达城市。cin >> start >> destination;cout << "Test Case " << cases++ << "." << endl;// 处理特殊情况：起始和目标城市相同。if (start == destination){cout << "Vladimir needs 0 litre(s) of blood." << endl;continue;}// 处理特殊情况：起始或目标城市不在输入数据中。if (cities.find(start) == cities.end()|| cities.find(destination) == cities.end()){cout << "There is no route Vladimir can take." << endl;continue;}// 取出起始和目标城市的序号。from = cities[start];to = cities[destination];// 遍历所有路线以找到从城市 from 到城市 to 所需的最少血量。int litres = travel(from, to);if (litres == NO_ROUTE)cout << "There is no route Vladimir can take." << endl;elsecout << "Vladimir needs " << litres << " litre(s) of blood." << endl;}return 0;}