Network Saboteur

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).
Output file must contain a single integer – the maximum traffic between the subnetworks.

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

Sample Output

90

本题的题意是将N台电脑分配到两个子网下面，子网内部没有距离，子网之间的电脑有距离，问怎样分子网使距离达到最大。
本题用深度搜索的方式枚举每一种分配方式，然后找最大的就行了。
代码如下：

#include <iostream>
#include <string.h>
#define MAXN 20 + 1
using namespace std;
int dis[MAXN][MAXN];
int a[MAXN];
int b[MAXN];
int flag[MAXN];
int n, max_num;
int an,bn;

void getDis()
{
    int d = 0;
    if(an == 0 || bn == 0)
        return;
    for(int i = 0; i < an; i++)
        for(int j = 0; j < bn; j++)
            d += dis[a[i]][b[j]];
    if(max_num < d)
        max_num = d;
}

void dfs(int i, int n)
{
    if(i > n)
    {
        bn = 0;
        for(int k = 1; k <= n; k++)
        {
            if(flag[k] == 0)
            {
                b[bn] = k;
                bn++;
            }
        }
        getDis();
    }
    else
    {
    //将第i台电脑加到一个子网中
        a[an] = i;
        an++;
        flag[i] = i;
        dfs(i+1,n);
        //将第i台电脑不加到这个子网，即加入另一个子网
        flag[i] = 0;
        a[an] = 0;
        an--;
        dfs(i+1,n);
    }
}

int main()
{
    while(cin >> n)
    {
        max_num = 0;
        an = 0, bn = 0;
        memset(dis, 0, sizeof(dis));
        for(int i = 1; i <= n; i++)
            for(int j = 1; j<=n; j++)
                cin >> dis[i][j];
        dfs(1,n);
        cout << max_num << endl;
    }
    return 0;
}