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 C1 C2 ……Cn

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

题目大意：

给出一条路径，判断这条路径是这个图的旅行商环路、简单旅行商环路还是非旅行商环路～

分析：

如果给出的路径存在某两个连续的点不可达或者第一个结点和最后一个结点不同或者这个路径没有访问过图中所有的点，那么它就不是一个旅行商环路(flag = 0)～如果满足了旅行商环路的条件，那么再判断这个旅行商环路是否是简单旅行商环路，即是否访问过n+1个结点（源点访问两次）～最后输出这些旅行商环路中经过的路径最短的路径编号和路径长度～

原文链接：https://blog.csdn.net/liuchuo/article/details/82560843

我的题解

感觉自己写的这堆shit逻辑很不清晰，本来拿了21分已经很不错了
结果发现minsum写的太小了，改了改惊奇地发现竟然AC了！震惊！！

#include <bits/stdc++.h>

using namespace std;
int n,m;
int e[210][210];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int a,b,dist;
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        cin>>a>>b>>dist;
        e[a][b]=e[b][a]=dist;
    }
    int k,num,id,minsum=999999;
    cin>>k;
    for(int query=1; query<=k; query++)
    {
        cin>>num;
        vector<int> v(num);
        set<int> s;
        int flag1=1,flag2=1,flag3=1;
        for(int i=0; i<num; i++)
        {
            cin>>v[i];
            s.insert(v[i]);
        }
        if(s.size()!=n||v[0]!=v[num-1])
            flag1=0;
        if(flag1)
        {
            if(num-1==n)
                flag2=0;
        }
        int sum=0;
        for(int i=0; i<num-1; i++)
        {
            if(e[v[i]][v[i+1]]==0)
            {
                flag3=0;
                flag1=0;
                flag2=0;
                break;
            }
            int dis=e[v[i]][v[i+1]];
            sum+=dis;
        }
        if(flag1&&!flag2)
        {
            if(sum<minsum)
            {
                id=query;
                minsum=sum;
            }
            printf("Path %d: %d (TS simple cycle)\n",query,sum);
        }
        else if(flag1&&flag2)
        {
            if(sum<minsum)
            {
                id=query;
                minsum=sum;
            }
            printf("Path %d: %d (TS cycle)\n",query,sum);
        }
        else if(!flag1&&flag3)
        {
            printf("Path %d: %d (Not a TS cycle)\n",query,sum);
        }
        else if(flag3==0)
        {
            printf("Path %d: NA (Not a TS cycle)\n",query);
        }
    }
    printf("Shortest Dist(%d) = %d",id,minsum);
    return 0;
}

柳神题解

还是学习一下大佬的题解吧Orz
重新理解了一下if...else if...else

隐藏的条件判断是：
条件2，默认条件1不成立；
条件3，默认条件1和条件2不成立；
...

#include <iostream>
#include <vector>
#include <set>
using namespace std;
int e[300][300], n, m, k, ans = 99999999, ansid;
vector<int> v;
void check(int index) {
    int sum = 0, cnt, flag = 1;
    scanf("%d", &cnt);
    set<int> s;
    vector<int> v(cnt);
    for (int i = 0; i < cnt; i++) {
        scanf("%d", &v[i]);
        s.insert(v[i]);
    }
    for (int i = 0; i < cnt - 1; i++) {
        if(e[v[i]][v[i+1]] == 0) flag = 0;
        sum += e[v[i]][v[i+1]];
    }
    if (flag == 0) {
        printf("Path %d: NA (Not a TS cycle)\n", index);
    } else if(v[0] != v[cnt-1] || s.size() != n) {
        printf("Path %d: %d (Not a TS cycle)\n", index, sum);
    } else if(cnt != n + 1) {
        printf("Path %d: %d (TS cycle)\n", index, sum);
        if (sum < ans) {
            ans = sum;
            ansid = index;
        }
    } else {
        printf("Path %d: %d (TS simple cycle)\n", index, sum);
        if (sum < ans) {
            ans = sum;
            ansid = index;
        }
    }
}
int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i++) {
        int t1, t2, t;
        scanf("%d%d%d", &t1, &t2, &t);
        e[t1][t2] = e[t2][t1] = t;
    }
    scanf("%d", &k);
    for (int i = 1; i <= k; i++) check(i);
    printf("Shortest Dist(%d) = %d\n", ansid, ans);
    return 0;
}