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
其實就這三句話，轉換成對應的判斷條件就好啦
- TS simple cycle if it is a simple cycle that visits every city; 如果是經過了所有結點並且是個簡單圓，經過所有結點好判斷，簡單圓判斷起來需要 路徑裏面的結點個數是所有結點+1，並且收尾結點一樣
- TS cycle if it is a cycle that visits every city, but not a simple cycle; 經過所有結點，並且首尾結點一樣
- Not a TS cycle if it is NOT a cycle that visits every city. 剩餘的情況
AC代碼
#include <iostream>
#include <string>
#include <vector>
#include <climits>
using namespace std;
struct node
{
string name;
int high;
};
int sort2(node a,node b )
{
if(a.high!=b.high)
return a.high>b.high;
else
return a.name<b.name;
}
int main()
{
int N,Ne;
cin>>N>>Ne;
int edge[N+1][N+1];
fill(edge[0],edge[0]+(N+1)*(N+1),-1);
for(int i=0; i<Ne; i++)
{
int a,b,c;
cin>>a>>b>>c;
edge[a][b]=c;
edge[b][a]=c;
}
int num;
cin>>num;
int cycleShort=INT_MAX;
int shortID;
for(int i=0; i<num; i++)
{
int pathLong;//記錄路徑個數
cin>>pathLong;
int sn[pathLong];
for(int i=0; i<pathLong; i++)
{
cin>>sn[i];
}
int nodeSign[N+1]= {0};
int pathOK=1;
int path=0;
for(int i=0; i<pathLong-1; i++)
{
if(edge[sn[i]][sn[i+1]]==-1)
{
pathOK=0;
break;
}
path+=edge[sn[i]][sn[i+1]];
nodeSign[sn[i]]=1;
nodeSign[sn[i+1]]=1;
}
if(pathOK==0)
{
printf("Path %d: NA (Not a TS cycle)",i+1);
}
else
{
//路徑肯定是沒問題了
//檢測是否訪問了所有結點
int allNodeSign=1;
for(int i=1; i<=N; i++)
{
if(nodeSign[i]==0)
{
allNodeSign=0;
break;
}
}
if(pathLong==N+1&&allNodeSign&&sn[0]==sn[pathLong-1]){
//簡單環並且訪問了所有結點
printf("Path %d: %d (TS simple cycle)",i+1,path);
if(path<cycleShort){
cycleShort=path;
shortID=i+1;
}
}
else if(allNodeSign&&sn[0]==sn[pathLong-1]){
printf("Path %d: %d (TS cycle)",i+1,path);
if(path<cycleShort){
cycleShort=path;
shortID=i+1;
}
}
else
printf("Path %d: %d (Not a TS cycle)",i+1,path);
}
cout<<endl;
}
printf("Shortest Dist(%d) = %d",shortID,cycleShort);
return 0;
}