pta 1150 Travelling Salesman Problem （25 分）（模拟）

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


题意：
给出n个城市的m条路径长度；
k组测试，p个点，如果这些点之间是不连通的，输出  NA (Not a TS cycle)  
如果连通成环，且遍历了所有城市，同时经过了n+1 个点（即没有重复经过某些城市） 输出   TS simple cycle
如果连通成环，且遍历了所有城市，但是重复经过了某些城市，输出  TS cycle

最后在这些成环的测试中选出最短的路径，并输出他是第几组测试数据，最短路径是多少。



我犯的错误：在找最短路径的时候忘记初始化，导致样例通过，提交上去测试点都没有通过。。。。。。
所以要注重细节


代码如下：

#include<bits/stdc++.h>
using namespace std;

int dist[202][202];//记录城市与城市之间的距离
int num[202];
int vis[202];//记录该城市有没有被访问过
int main()
{
    int n,m,p,a,b,d,k,count,cnt,ans,ansid;
    cin >> n >> m;
    memset(dist,-1,sizeof(dist));//初始化为-1
    while(m--)//初始化
    {
        cin >> a >> b >> d;
        dist[a][b] = d;
        dist[b][a] = d;
    }
    cin >> k;
    ans = 0x3f3f3f3f;//初始化
    for(int i = 1;i <= k;i++)
    {
        count = 0,cnt = 0;//初始化路径总长度和经过的城市个数
        cin >> p;
        memset(vis,0,sizeof(vis));
        for(int j = 0;j < p;j++)
        {
            cin >> num[j];
            if(vis[num[j]] == 0)//没有被访问过
            {
                vis[num[j]] = 1;
                cnt ++;
            }
        }
        cout << "Path " << i << ": ";
        int j;
        for(j = 0;j < p - 1;j++)
        {
            if(dist[num[j]][num[j + 1]] == -1)//不连通
            {
                cout << "NA (Not a TS cycle)" << endl;
                break;
            }
            else
            count += dist[num[j]][num[j + 1]];
        }
        if(j == p - 1)//连通
        {
            cout << count << " ";
            //cout << cnt << "-----" << num[0]  << "---" << num[p - 1] << endl;
            if(cnt == n && num[0] == num[p - 1])//遍历了所有节点，且成环
            {
                if(count < ans)//寻找最短路
                {
                    ans = count;
                    ansid = i;
                }
                if(p == n + 1)//没有经过重复的城市
                cout << "(TS simple cycle)" << endl;
                else
                cout << "(TS cycle)" << endl;
            }
            else
            cout << "(Not a TS cycle)" << endl;
        }
    }
    cout << "Shortest Dist(" << ansid << ") = " << ans << endl; 
    return 0;
}