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POJ 2531 Network Saboteur(搜索)


Network Saboteur


Time Limit: 2000MS

 

Memory Limit: 65536K

Total Submissions: 10084

 

Accepted: 4819


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


Source


Northeastern Europe 2002, Far-Eastern Subregion






      题意:将n个点分成两个集合,同集合内的点没有距离,不同集合内的点有距离,求出他们分成两个集合后所能达到的最大距离




点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<vector>

using namespace std;

int map[25][25];
int v[25];
int n;
int maxx;

int Num()
{
    int sum = 0;
    for(int i=0; i<n; i++)
    {
        if(v[i] == 1)
        {
            for(int j=0;j<n;j++)
            {
                if(v[j]!=1)
                {
                    sum += map[i][j];
                }
            }
        }
    }
    return sum;
}

void DFS(int m)
{
    if(m == n)
    {
         int sum = Num();
         maxx = max(maxx,sum);
         return ;
    }
    for(int i=0; i<2; i++)
    {
        v[m] = i;
        DFS(m+1);
    }
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                scanf("%d",&map[i][j]);
            }
        }
        memset(v,0,sizeof(v));
        maxx = 0;
        DFS(0);
        printf("%d\n",maxx);
    }
    return 0;
}




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