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<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

不难就是麻烦

//邻接矩阵存图，如果这条路径上的点都只出现一次，就是simple的，如果重复出现，就不是simple，如果前后不同或者中间断开另外讨论
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std;
const int maxn = 205;
int G[maxn][maxn];int main(){int n,m;int minDis = 1000000000,ans;cin>>n>>m;int u,v,dis;fill(G[0], G[0]+maxn*maxn,0);for(int i = 0; i < m; i++){cin>>u>>v>>dis;G[u][v] = G[v][u] = dis;}int k, num,x;cin>>k;for(int i = 1; i <= k; i++){cin>>num;int hashTable[maxn] = {0}, tempDis = 0;bool cycle = true;bool haveDis = true;bool simpleCycle = true;vector<int> path;for(int j = 0; j < num; j++){cin>>x;path.push_back(x);}if(path[0] != path[num-1]) cycle = false;for(int j = 0; j < num-1; j++){hashTable[path[j]]++;if(G[path[j]][path[j+1]] != 0) tempDis+=G[path[j]][path[j+1]];else haveDis = false;}for(int j = 1; j <= n; j++){if(j!=path[0] && hashTable[j] >=2) simpleCycle = false;}//还有一个判断circle就是是不是所有点都被访问if(num < n){cycle = false;simpleCycle = false;}for(int j = 1; j <= n; j++){if(hashTable[j] == 0) cycle = false;}if(tempDis < minDis && haveDis == true && cycle == true){minDis = tempDis;ans = i;}if(cycle == true && simpleCycle == true && haveDis==true) printf("Path %d: %d (TS simple cycle)\n",i,tempDis);else if(cycle == true && simpleCycle == false &&haveDis == true) printf("Path %d: %d (TS cycle)\n",i,tempDis);else if(cycle == false && haveDis == true) printf("Path %d: %d (Not a TS cycle)\n",i,tempDis);else printf("Path %d: NA (Not a TS cycle)\n",i);}printf("Shortest Dist(%d) = %d",ans,minDis);return 0;
}

二刷，写的差不多

//simple cycle:最后一个等于第一个，路径是通路，点集个数 -1等于n，去重后的个数=n
//ts cycle:最后一个等于第一个，路径通路，点集个数-1不等于n
//not:最后一个字符不等于第一个或者中间不连通
#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<set>
using namespace std;
const int maxn = 205;
int G[maxn][maxn];
int main(){fill(G[0],G[0]+maxn*maxn,0);int n,m;cin>>n>>m;int u,v,dis;int minDis = 1000000000;int res;for(int i = 0; i < m; i++){cin>>u>>v>>dis;G[u][v] = G[v][u] = dis;}int q,k,dian;cin>>q;for(int i = 1; i <= q; i++){cin>>k;vector<int> v;set<int> s;int distance = 0;bool connect = true;bool simple = true;bool cycle = true;for(int j = 0; j < k; j++){cin>>dian;v.push_back(dian);s.insert(dian);}for(int j = 0; j < k - 1; j++){if(G[v[j]][v[j+1]] != 0){distance += G[v[j]][v[j+1]];}else{connect = false;break;}}if(k - 1 != n) simple = false;if(v[k-1] != v[0] || s.size() != n) cycle = false;if(simple == true && connect == true && cycle == true) printf("Path %d: %d (TS simple cycle)\n",i,distance);else if(simple == false && connect == true && cycle == true) printf("Path %d: %d (TS cycle)\n",i,distance);else if(cycle == false && connect == true) printf("Path %d: %d (Not a TS cycle)\n",i,distance);else if(connect == false) printf("Path %d: NA (Not a TS cycle)\n",i);if(connect == true && cycle == true){if(distance < minDis){minDis = distance;res = i;}}}printf("Shortest Dist(%d) = %d",res,minDis);return 0;
}

没有全访问也不是圆，不是判断j和j+1有没有连通，而是v[j]和v[j+1]

//simple cycle:访问了每一个点，并且最后一个点和原点相等，用set来判断是否访问了每一个点
//ts cycle:访问了每一个点，最后一个点和原点相等，但是k-1和n不相等
//not cycle:最后一个点和原点不相等或者没有访问每一个点；两种情况，一种是通的，求出距离，一种是断的，na
#include<iostream>
#include<vector>
#include<unordered_set>
#include<cstdio>
using namespace std;
const int maxn = 205;
int G[maxn][maxn];int main(){int n,m;cin>>n>>m;int u,v,dis;for(int i = 0; i < m; i++){cin>>u>>v>>dis;G[u][v] = G[v][u] = dis;}int q,k;cin>>q;int minDis = 1e9;int ans;for(int i = 1; i <= q; i++){cin>>k;vector<int> v(k+1);unordered_set<int> s;int totalDis = 0;bool isCycle = true;//有两种情况会让它不是圆bool connected = true;bool isSimple = true;for(int j = 0; j < k; j++){cin>>v[j];s.insert(v[j]);}if(v[0] != v[k-1]) isCycle = false;for(int j = 0; j < k-1; j++){if(G[v[j]][v[j+1]] == 0){connected = false;isCycle = false;break;}totalDis += G[v[j]][v[j+1]];}if(k-1 != n) isSimple = false;if(s.size() != n) isCycle = false;if(connected == false) printf("Path %d: NA (Not a TS cycle)\n",i);else if(isCycle == false && connected == true) printf("Path %d: %d (Not a TS cycle)\n",i,totalDis);else if(isSimple == false && isCycle == true) printf("Path %d: %d (TS cycle)\n",i,totalDis);else if(isSimple == true && isCycle == true) printf("Path %d: %d (TS simple cycle)\n",i,totalDis);if(isCycle == true && totalDis < minDis){minDis = totalDis;ans = i;}}printf("Shortest Dist(%d) = %d\n",ans,minDis);return 0;
}

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