C language using backtracking method to solve the problem of traveling salesman and the problem of M coloring of graphs _c language

Travel salesman problem
1. Question Description:

The problem of traveling salesman is also called TSP. The problem is as follows: A salesman to a number of cities to sell goods, known to the distance between the city (or travel), he would like to select a departure from the station, through each city the final return to the station route, so that the overall route (or total travel) the smallest. The mathematical model for given a graph, to traverse each vertex once and only once a loop, finally back to the minimum cost of starting point.

2. Input requirements:

The first behavior of the input test sample number T (T < 120), followed by a test sample of T. The first row of each test sample is the number of vertices n, the number of edges m (N < 12,m < 100) of the graph without direction, and then the M row, three integers u, V, and W in each row, indicating that a side of the vertex U and V has a weighted value of W. (1 <= u < v <= n,w <= 1000). Assume the starting point (the station) is the vertex of number 1th.

3. Output Requirements:

For each test sample output line, the format is "Case #: W", where ' # ' indicates the first few test samples (starting from 1), W is the optimal solution of TSP problem, if the feasible solution can not find the output-1.

4. Sample input:

2
5 8
1 2 5
1
4
7 1 5 9 2 3 2 4 3 2 5 6 3 4 8 4 5 4-3 1 1 2

5. Sample output:

Case 1:36 Case
2:-1

6. Solution:

Travel salesman problem (backtracking) #include <iostream> #define N-namespace Std; int n,m,w,//graph vertex number and edge number graph[n][n],//graph weighted adjacency matrix c=0,//Current cost bestc=-1,//Current optimal value x[n],//current solution be    Stx[n]; 
The current optimal solution void backtrack (int k); 
void swap (int &a,int &b); 
  void swap (int &a,int &b) {int temp=a; 
  A=b; 
B=temp; } void Backtrack (int k) {if (k==n) {if (c+graph[x[n-1]][x[n]]+graph[x[n]][1]<bestc| | Bestc==-1) && graph[x[n-1]][x[n]]!=-1 && graph[x[n]][1]!=-1) {bestc=c+graph[x[n-1]][x[n]]+gr 
      APH[X[N]][1]; 
      for (int i=1;i<=n;i++) {bestx[i]=x[i]; 
  } return; else {for (int i=k;i<=n;i++) {if graph[x[k-1]][x[i]]!=-1 && C+GRAPH[X[K-1]][X[I]]&L T;BESTC | | 
        Bestc==-1)) {swap (x[i],x[k]); 
        C+=graph[x[k-1]][x[k]]; 
        Backtrack (k+1); 
        C-=graph[x[k-1]][x[k]]; 
      Swap (x[i],x[k]); 
} 
    }  int main (void) {int i,j,tmp=1,testnum;
  cin>>testnum;
    while (Tmp<=testnum) {cin>>n>>m;
    for (i=1;i<=n;i++) for (j=1;j<=n;j++) graph[i][j]=-1;
      for (int k=1;k<=m;k++) {cin>>i>>j>>w;
      Graph[i][j]=w;
    Graph[j][i]=w;
      for (i=1;i<=n;i++) {x[i]=i;
    Bestx[i]=i;
    } backtrack (2);
    cout<< "Case" <<tmp<< ":" <<bestc<<endl;
    Bestc=-1;
    
    c=0;
  tmp++;
return 0;

 }

The problem of M coloring of graphs
1. Description of the problem
given undirected connected graphs G and m are different colors. These colors are colored for each vertex of fig g, each vertex a color. Whether there is a coloring method of g in each side of the 2 vertices of different colors, ask how many methods for the figure can be m coloring.

2. Input requirements:
the first input is the number T (T < 120) of the test sample, followed by a test sample of T. The first line of each test sample is the number of vertices n, the number of edges m and the number of available colors m (n <= 10,m < 100,m <= 7), followed by m rows, two integers u and V in each row, representing a side connection between the vertices U and v. (1 <= u < v <= N).

3. Output Requirements:
for each test sample output two lines, the first line is "Case #: W", where ' # ' indicates the first few test samples (starting from 1), and W is the number of M-shaded schemes.

4. Sample input:

1
5 8 5
1 2
1
3
1 4 2 3 2 4 2 5 3 4 4-5

5. Sample output:

Case 1:360

6. Solution:

#include <iostream> using namespace std;
#define N-M,n,m,a[n][n],x[n],textnum int;

int static sum=0;
  bool OK (int k) {for (int j=1;j<=n;j++) if (a[k][j]&& (x[j]==x[k)) is return false;
return true;
    } void Backtrack (int t) {if (t>n) {sum++;
    for (int i=1;i<=n;i++)//cout<<x[i]<< "";
  cout<<endl;
    else for (int i=1;i<=m;i++) {x[t]=i;
    if (OK (t)) backtrack (t+1);
  x[t]=0;
  int main () {int i,j,z=1;         cin>>textnum;          Enter the number of tests while (textnum>0) {cin>>n;
    The number of input vertices for (i=1;i<=n;i++) for (j=1;j<=n;j++) a[i][j]=0;         cin>>m>>m;
      Number of input edges, number of available colors for (int k=1;k<=m;k++)//graph adjacency Matrix {cin>>i>>j;
      A[i][j]=1;
    A[j][i]=1;
    }/* for (i=1;i<=n;i++) {for (j=1;j<=n;j++) cout<<a[i][j]<< "; Cout<<endl;}
    * * for (i=0;i<=n;i++) x[i]=0; BaCktrack (1);
    cout<< "Case" <<z<< ":" <<sum<<endl;
        
    sum=0;
    textnum--;
  z++;
return 0;
 }