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Network Saboteur

2022-08-10 13:18:00 51CTO



Problem Description


A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully divided the network into two subnetworks in order to minimize traffic between parts.
A disgruntled computer science student Vasya, after being expelled from the university, decided to have his revenge. He hacked into the university network and decided to reassign computers to maximize the traffic between two subnetworks.
Unfortunately, he found that calculating such worst subdivision is one of those problems he, being a student, failed to solve. So he asks you, a more successful CS student, to help him.
The traffic data are given in the form of matrix C, where Cij is the amount of data sent between ith and jth nodes (Cij = Cji, Cii = 0). The goal is to divide the network nodes into the two disjointed subsets A and B so as to maximize the sum ∑Cij (i∈A,j∈B).


Input


The first line of input contains a number of nodes N (2 <= N <= 20). The following N lines, containing N space-separated integers each, represent the traffic matrix C (0 <= Cij <= 10000). <br>Output file must contain a single integer -- the maximum traffic between the subnetworks. <br>


Output


Output must contain a single integer -- the maximum traffic between the subnetworks.


Sample Input



      
      
3
0 50 30
50 0 40
30 40 0
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Sample Output


      
      
90
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Many points are directly distant,Divide the points into two-sum sets,The division method of the largest of the two sets.


思路:


用dfs()Repeatedly divide the points into two sets.


代码:


      
      
#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int n;
int map[ 22][ 22];
int po[ 22];
int sum = 0;
int sun;
int dfs( int k)
{
if( k == n + 1)
{
if( sum < sun) sum = sun;
return 0;

}


int an = 0;
for( int j = 1; j <= k - 1; j ++)
{
if( po[ j]){
an += map[ k][ j];

}
}
sun += an;
dfs( k + 1);
sun -= an;
po[ k] = 1;
an = 0;
for( int j = 1; j <= k - 1; j ++)
{
if( ! po[ j]){
an += map[ k][ j];

}
}
sun += an;
dfs( k + 1);
sun -= an;
po[ k] = 0;



}

int main()
{
while( cin >> n)
{ memset( map, 0, sizeof( map));
memset( po, 0, sizeof( po));
sum = 0;
sun = 0;
for( int i = 1; i <= n; i ++)
{
for( int j = 1; j <= n; j ++)
{
scanf( "%d", & map[ i][ j]);
}
}

po[ 1] = 1;
dfs( 1);
printf( "%d\n", sum);
}


return 0;
}
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