Traveling Salesman Problem (depth-first search)

Input: An integer n, n * n Then there rectangle, representing the distance between the two cities to spend, spend between each two cities no more than 10,000.

Output: an integer representing the No. 1 tourist attractions all over the city and back to the minimum cost No. 1 of the city.

Sample input:

4

0 1 1 1

1 0 2 1

5 5 0 6

1 1 3 0

Sample output:

8

Thinking:

1. End search all cities, which represents a depth-first search, coupled with the return from the end to take the No. 1 city that is a full cost fees. So every search deep end (End search all cities) represents a cost results. Record minimum cost to all deep in search of.

2 marks each process has reached deep search in the city, the city has not yet reached to continue the search, the search after the end of a deep, clear the mark. Deep search for the next, resulting in different costs result.

For example: 1,2,3,4 four cities.

Search options:

1->2->3->4、1->2->4->3、1->3>2->4、1->3->4->2、1->4->2->3、1->4->3->2 ......

. 1 #include <the iostream>
 2 #include <cstdio>
 . 3  the using  namespace STD;
 . 4  int n-;
 . 5  int ANS = 100000000 ;
 . 6  int G [ 100 ] [ 100 ];
 . 7  BOOL VIS [ 100 ] = { 0 };
 . 8  void the DFS ( int U, int cnt, int sum) {
 9      // optimality pruning: in a deep search process cost> current completed all the required sum from the cost of ANS,
 10      // therefore certainly not a minimum sum , not in the search to the lower, the function returns continue to look for another solution. 
11     IF (SUM> ANS) {
 12 is          return ;
 13 is      }
 14      IF (CNT == n-) {
 15          ANS = min (ANS, SUM + G [U] [ . 1 ]);
 16      }
 . 17      VIS [U] = to true ; / / marker present layer recursion searched 
18 is      for ( int I = . 1 ; I <= n-; I ++ ) {
 . 19          IF (! {VIS [I])
 20 is              DFS (I, CNT + . 1 , SUM + G [U] [I] );
 21 is          }
 22 is      }
 23 is      VIS [U] =to false ; // clear the flag after the completion of a deep search 
24  }
 25  int main ()
 26 is  {
 27      CIN >> n-;
 28      for ( int I = . 1 ; I <= n-; I ++ ) {
 29          for ( int J = . 1 ; J <= n-; J ++ ) {
 30              CIN >> G [I] [J];
 31 is          }
 32      }
 33 is      DFS ( . 1 , . 1 , 0 );
 34 is      COUT ANS << << endl;
 35      return 0;
36 }