1150 Travelling Salesman Problem （25 分) 模拟

淡淡的烟草味﹌ 2022-02-26 02:56 224阅读 0赞

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][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 Ci'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

居然不用最短路，直接把给的边存下来输出就行了...

判断三种情况，1简单的环，每个点出现一次，循环一遍。

2每个点都访问，但是不止一次

3不是环（其中包括收尾不连，有不可达点）

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<sstream>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    int ma[205][205],s[1005];
    int main()
    {
    	int n,m,k,i,j,l,x,y,z;
    	scanf("%d%d",&n,&m);
    	memset(ma,inf,sizeof ma);
    	for(i=0;i<m;i++)
    	{
    		scanf("%d%d%d",&x,&y,&z);
    		if(ma[x][y]>z)
    		ma[x][y]=ma[y][x]=z;
    	}
    	scanf("%d",&m);
    	int o=1,max=inf,maxi;
    	while(m--)
    	{
    		int v[205]={0};
    		scanf("%d",&k);
    		for(i=0;i<k;i++)
    		{
    			scanf("%d",&s[i]);
    			v[s[i]]++;
    		}		
    		printf("Path %d: ",o++);
    		int sum=0,f=0;
    		for(i=0;i<k-1;i++)
    		{
    			//printf("**%d %d %d\n",s[i],s[i+1],ma[s[i]][s[i+1]]);
    			if(ma[s[i]][s[i+1]]!=inf)
    			sum+=ma[s[i]][s[i+1]];
    			else
    			f=1;
    		}
    		if(f==0)
    		printf("%d",sum);	
    		else
    		printf("NA");	
    		if(f==1||s[0]!=s[k-1]||k<=n)//有不可达点or收尾不相同or结点数不足以循环 
    		printf(" (Not a TS cycle)\n");
    		else
    		{
    			if(sum<max)
    			{
    				max=sum;
    				maxi=o;
    			}
    			int f0=0,f1=0,f2=0;
    			for(i=1;i<=n;i++)
    			{
    				if(v[i]==0)f0++;
    				else if(v[i]==1)f1++;
    				else if(v[i]==2)f2++;
    			}
    			if(f2==1&&f1==n-1)//仅有收尾节点被访问2次 
    			printf(" (TS simple cycle)\n");
    			else if(f0==0)//每个结点都被访问到了 
    			printf(" (TS cycle)\n");
    			else
    			printf(" (Not a TS cycle)\n");
    		}
    	}
    	printf("Shortest Dist(%d) = %d\n",maxi-1,max);
    }

[https_en.wikipedia.org_wiki_Travelling_salesman_problem]: https://en.wikipedia.org/wiki/Travelling_salesman_problem