把一个完全图分成两部分

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

//Memory Time
//188K   375MS 

#include<iostream>
using namespace std;

const int TimeLimit=2000;  //本题时间限制为2000ms

int main(int i,int j)
{
    int n;
    while(cin>>n)
    {
        /*Input*/

        int w[30][30]={0};
        for(i=1;i<=n;i++)
            for(j=1;j<=n;j++)
            {
                cin>>w[i][j];
                w[j][i]=w[i][j];  //双向完全图
            }

        /*Random Algorithm*/

        bool subset[30]={false};    //A集:true  B集:false
        int time=TimeLimit*100;  //使随机次数尽可能大，随机结果尽可能接近最优解
        long max_w=0;   //最大割的权值之和
        long sum=0;  //当前边割集权和

        while(time--)
        {
            int x=rand()%n+1;  //生成随机数 x，对应于总集合的某个结点x
                               //注意由于使用的结点序号为1~n，对应了数组下标，下标为0的数组元素没有使用
                               //那么这里必须+1，因为若rand()=n，那么再对n取模结果就为0
                               //这时就会导致使用了不存在的 [0]结点，本应使用的 [n]结点就被丢弃了

            subset[x]=!subset[x];  //改变x所在的集合位置

            for(int i=1;i<=n;i++)   //由于是完全图，所以每个顶点i都与x相关联，因此要全部枚举
            {
                if(subset[i]!=subset[x])   //结点i 和 x分别在两个集合内
                    sum+=w[i][x];   //就是说因为x所在集合的改变，使得割边的个数增加
                                    //割集的原权值 要加上 当前新加入的割边(i,x)的权值

                if(i!=x && subset[i]==subset[x])  //结点i 和 x分别在相同的集合内，但他们不是同一元素
                    sum-=w[i][x];   //就是说因为x所在集合的改变，使得割边的个数减少
                                    //割集的原权值 要减去 当前失去的割边(i,x)的权值
            }

            if(max_w < sum)
                max_w = sum;
        }

        cout<<max_w<<endl;
    }
    return 0;
}