PAT A1150 Travelling Salesman Problem （25 分）——图的遍历

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 C​1​​ C​2​​ ... C​n​​

where n is the number of cities in the list, and C​i​​'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

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <map>
#include <vector>
#include <set>
using namespace std;
int n,m,k;
int dis[][];
int path[],vis[];
int main(){
    scanf("%d %d",&n,&m);
    for(int i=;i<m;i++){
        int c1,c2,d;
        scanf("%d %d %d",&c1,&c2,&d);
        dis[c1][c2]=d;
        dis[c2][c1]=d;
    }
    scanf("%d",&k);
    int min=,mini=;
    for(int i=;i<=k;i++){
        int flag=;
        int total=;
        int nn;
        fill(vis,vis+,);
        scanf("%d",&nn);
        for(int j=;j<nn;j++){
             scanf("%d",&path[j]);
             vis[path[j]]++;
        }
        for(int j=;j<=n;j++){
             if(vis[j]==) flag=;
        }
        for(int j=;j<nn;j++){
              if(dis[path[j]][path[j-]]==){
                  total=-;
                  flag=;
                  break;
            }
            else{
                total+=dis[path[j]][path[j-]];
            }
          }
          printf("Path %d: ",i);
          if(total==-) printf("NA ");
          else printf("%d ",total);
          if(flag== || path[]!=path[nn-]) printf("(Not a TS cycle)\n");
          else{
              if(total<min){
                  min=total;
                  mini=i;
            }
              if(nn==n+) printf("(TS simple cycle)\n");
              else printf("(TS cycle)\n");
        }
    }
    printf("Shortest Dist(%d) = %d\n",mini,min);
}

注意点：看到题目一直不知道怎么做，这似乎是一个从一个点出发，找到最短的回到原点的路径，又不是最小生成树，也不是全源最短路径。没有一个已知算法适合做这个。没办法只好看大神思路，看了以后发现什么鬼，

TS simple cycle 居然是判断给定路径是不是都遍历了所有城市，并且起点和终点相同，只有起点重复了一次，只是看是否是最简单的环，并不管路径长度

TS cycle 就是判断给定路径是不是遍历了所有城市，但不是最简单的环，即有城市访问太多遍了

Not a TS cycle 是看给定路径有没有到所有城市，起点终点一不一样，有没有路走不通的