PAT A1150 Travelling Salesman Problem (25point(s))

Scroll Down

The "travelling salesman problem" asks the following question: "Given a list of cities and the distances between each pair of cities, what is the shortest possible route that visits each city and returns to the origin city?" It is an NP-hard problem in combinatorial optimization, important in operations research and theoretical computer science. (Quoted from "https://en.wikipedia.org/wiki/Travelling_salesman_problem".)

In this problem, you are supposed to find, from a given list of cycles, the one that is the closest to the solution of a travelling salesman problem.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers N (2<N≤200), the number of cities, and M, the number of edges in an undirected graph. Then M lines follow, each describes an edge in the format City1 City2 Dist, where the cities are numbered from 1 to N and the distance Dist is positive and is no more than 100. The next line gives a positive integer K which is the number of paths, followed by K lines of paths, each in the format:
n C1C2...Cn

where n is the number of cities in the list, and Ci's are the cities on a path.

Output Specification:

For each path, print in a line Path X: TotalDist (Description) where X is the index (starting from 1) of that path, TotalDist its total distance (if this distance does not exist, output NA instead), and Description is one of the following:

TS simple cycle if it is a simple cycle that visits every city;
TS cycle if it is a cycle that visits every city, but not a simple cycle;
Not a TS cycle if it is NOT a cycle that visits every city.
Finally print in a line Shortest Dist(X) = TotalDist where X is the index of the cycle that is the closest to the solution of a travelling salesman problem, and TotalDist is its total distance. It is guaranteed that such a solution is unique.

Sample Input:

6 10
6 2 1
3 4 1
1 5 1
2 5 1
3 1 8
4 1 6
1 6 1
6 3 1
1 2 1
4 5 1
7
7 5 1 4 3 6 2 5
7 6 1 3 4 5 2 6
6 5 1 4 3 6 2
9 6 2 1 6 3 4 5 2 6
4 1 2 5 1
7 6 1 2 5 4 3 1
7 6 3 2 5 4 1 6

Sample Output:

Path 1: 11 (TS simple cycle)
Path 2: 13 (TS simple cycle)
Path 3: 10 (Not a TS cycle)
Path 4: 8 (TS cycle)
Path 5: 3 (Not a TS cycle)
Path 6: 13 (Not a TS cycle)
Path 7: NA (Not a TS cycle)
Shortest Dist(4) = 8

题意

旅行家问题,给出一条路径,判断是否恰好经过所有城市以及是不是环。

解答

用set存储经过的城市序号,如果集合大小恰好为城市的总数,则经过了所有城市,如果经过的城市数目为n+1,第一个城市和最后一个城市一样,则就是TS simple cycle,如果经过的城市数目大于n+1,则是TS cycle,其他情况为Not a TS cycle。
注意:不是所有城市都邻接,因此要判断路线中是否有不可达的点。

#include<iostream>
#include<vector>
#include<set>

using namespace std;

int main()
{
	int citiesN, edgesN, a, b, cost;
	cin >> citiesN >> edgesN;
	vector<vector<int>> adj(citiesN + 1, vector<int>(citiesN + 1, -1));
	for (int i = 0; i < edgesN; i++) {
		cin >> a >> b >> cost;
		adj[a][b] = adj[b][a] = cost;
	}
	int pathN, minDistance = 100000, min = 1;
	cin >> pathN;
	for (int i = 0; i < pathN; i++) {
		int n, tmp, source, start, distance = 0;
		set<int> cities;
		cin >> n >> source;
		start = source;
		for (int j = 1; j < n; j++) {
			cin >> tmp;
			cities.insert(tmp);
			// 非邻接
			if (adj[source][tmp] < 0) 
				distance = -1;
			else if (distance >= 0) 
				distance += adj[source][tmp];
			source = tmp;
		}
		if (distance > 0 && cities.size() == citiesN && start == tmp) {
			if (n == citiesN + 1) printf("Path %d: %d (TS simple cycle)\n", i + 1, distance);
			else printf("Path %d: %d (TS cycle)\n", i + 1, distance);
			if (distance > 0 && distance < minDistance) {
				minDistance = distance;
				min = i + 1;
			}
		}
		else if (distance > 0)
			printf("Path %d: %d (Not a TS cycle)\n", i + 1, distance);
		else 
			printf("Path %d: NA (Not a TS cycle)\n", i + 1);
	}
	printf("Shortest Dist(%d) = %d", min, minDistance);
	return 0;
}