1150 Travelling Salesman Problem（25 分）

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), 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
给出一个无向有权图，然后给出k个路径，进行判断，是否是能访问所有城市的简单环，显然需要记录访问了几个城市，以及路径是否通，如果路径不通直接是NA，然后考虑其他的，要形成环，路径最少得有n + 1个点，且首尾要相同，而且路径要访问所有点，如果都满足了，要判断是不是简单环，简单环必须是n + 1个点。
代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#define inf 0x3f3f3f3f
#define MAX
using namespace std;
int mp[201][201];
int path[400];
int n,m,k;
int u,v,w,kk,mint,mind = inf;
int main() {
    scanf("%d%d",&n,&m);
    for(int i = 0;i < m;i ++) {
        scanf("%d%d%d",&u,&v,&w);
        mp[u][v] = mp[v][u] = w;
    }
    scanf("%d",&k);
    for(int i = 1;i <= k;i ++) {
        scanf("%d",&kk);
        int vis[201] = {0},c = 0,d = 0;
        for(int j = 0;j < kk;j ++) {
            scanf("%d",&path[j]);
            if(!vis[path[j]]) c ++;
            vis[path[j]] ++;
        }
        for(int j = 1;j < kk;j ++) {
            if(mp[path[j]][path[j - 1]]) {
                d += mp[path[j]][path[j - 1]];
            }
            else {
                c = -1;
                break;
            }
        }
        if(c == -1) printf("Path %d: NA (Not a TS cycle)
",i);
        else if(kk <= n || c < n || path[0] != path[kk - 1]) printf("Path %d: %d (Not a TS cycle)
",i,d);
        else {
            if(kk == n + 1) printf("Path %d: %d (TS simple cycle)
",i,d);
            else printf("Path %d: %d (TS cycle)
",i,d);
            if(mind > d) {
                mint = i;
                mind = d;
            }
        }
    }
    printf("Shortest Dist(%d) = %d",mint,mind);
}