10个图遍历的问题及答案，有兴趣的看看

2023年8月24日 343次阅读来源: 数据结构之图

PROBLEM : Star Gates

Problem: The Intelligent CargoExpress transport company of WorldZewa Planet
Allies (WZPA) services a number of planets in World Zewa. Because of
monetary concerns, all of the cargo space ships are ‘one-way.’ That is, a
route from Planet Kaitaia to Planet Invercargill does not imply the
existence of a route from Planet Invercargill to Planet Kaitaia. In fact,
even if both of these routes do happen to exist, they are distinct and are
not necessarily the same distance! The purpose of this problem is to help
the cargo transport company provide its customers with information about the
routes. In particular, you will compute the distance along a certain route,
the number of different routes between two planets, and the shortest route
between two planets.

Input: A directed graph where a node represents a planet and an edge
represents a route between two planets. The weighting of the edge
represents the distance between the two planets. A given route will never
appear more than once, and for a given route, the starting and ending planet
will not be the same planet.

Output: For test input 1 through 5, if no such route exists, output ‘NO SUCH
ROUTE’. Otherwise, follow the route as given; do not make any extra stops!
For example, the first problem means to start at planet A, then travel
directly to planet B (a distance of 5), then directly to planet C (a
distance of 4). /n /n
(Read from stdin, fixed in code also acceptable. No need to validate.) /n
AB5, BC4, CD8, DC8, DE6, AD5, CE2, EB3, AE7 /n

Expected Output:

Output #1: 9
Output #2: 5
Output #3: 13
Output #4: 22
Output #5: NO SUCH ROUTE
Output #6: 2
Output #7: 3
Output #8: 9
/nOutput #9: 9
Output #10: 7
/n
“,1] ); //–>

1. The distance of the route A-B-C.

2. The distance of the route A-D.

3. The distance of the route A-D-C.

4. The distance of the route A-E-B-C-D.

5. The distance of the route A-E-D.

6. The number of trips starting at C and ending at C with a maximum of 3

stops. In the sample data below, there are two such trips: C-D-C (2 stops).

and C-E-B-C (3 stops).

7. The number of trips starting at A and ending at C with exactly 4 stops.

In the sample data below, there are three such trips: A to C (via B,C,D); A

to C (via D,C,D); and A to C (via D,E,B).

8. The length of the shortest route (in terms of distance to travel) from A

to C.

9. The length of the shortest route (in terms of distance to travel) from B

to B.

10. The number of different routes from C to C with a distance of less than

30. In the sample data, the trips are: CDC, CEBC, CEBCDC, CDCEBC,

CDEBC,CEBCEBC, CEBCEBCEBC.

Test Input:

For the test input, the planets are named using A to Z. A route between two

planets (A to B) with a distance of 5 is represented as AB5.

Graph:
(Read from stdin, fixed in code also acceptable. No need to validate.)
AB5, BC4, CD8, DC8, DE6, AD5, CE2, EB3, AE7

Expected Output:

Output #1: 9

Output #2: 5

Output #3: 13

Output #4: 22

Output #5: NO SUCH ROUTE

Output #6: 2

Output #7: 3

Output #8: 9

Output #9: 9

Output #10: 7 ———————————————————————————————————– package cxg2; import java.util.Map; public interface ProblemState { public long processProblem(Map map);

}

————————————————————————— package cxg2; import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.ArrayList;

import java.util.HashMap;

import java.util.Iterator;

import java.util.List;

import java.util.Map; public class Method1 {

//存储权重状态

private Map powerStateMap = new HashMap();

//存储显示状态

private List viewStateList = new ArrayList();

//某个状态的输出结果

private long saveOutPut ;

//用于输出显示

private String printOutPut = null;

public static void main(String[] args){

Method1 instance = new Method1();

instance.getInput();

instance.processProblemState();

instance.viewPrint();

}

* 从控制台获得输入

private void getInput(){

try {

BufferedReader stdIn = new BufferedReader(

new InputStreamReader(System.in));

String userInput;

userInput = stdIn.readLine();

System.out.println(“Expected Output:”);

processInput(userInput);

stdIn.close();

}catch(IOException e){

System.err.println(“Couldn’t get I/O .”);

System.exit(1);

}

* 对输入的字符串处理

* 注意输入时请去处空格

private void processInput(String input){

String[] powers = input.split(“,”);

for(int i=0;i<powers.length;i++){

insertStateMap(powers[i]);

}

* 对每个输入的权重存入状态Map中

private void insertStateMap(String power){

//假设这里输入的都是不同的,AB5,BC4 存为{ key:AB value:5 }

String route = power.substring(0,2);

String powerSize = power.substring(2);

powerStateMap.put(route,powerSize);

}

* 利用反射机制可以减少代码

* 如果修改修改前5个state应该可以合并

private void processProblemState(){

//1

ProblemState state1 = new ConcreteProblemState1();

saveOutPut = state1.processProblem(powerStateMap);