Network Saboteur (poj 2531 dfs)

Language: Default

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9379		Accepted: 4430

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：给一个无向图，把它分成两部分，使得连接这两部分边的权和最大。

思路：定义一个group数组标记每个点的分组，从0号点dfs直到N，暴搜所有情况求最大值。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;

int mp[25][25];
int group[25];//标记分组
int N,ans;

void dfs(int d,int sum)
{
    if (d==N)
    {
        if (sum>ans)
            ans=sum;
        return ;
    }
    int all=0;
    for (int i=0;i!=d;i++)
        if (group[i]==1)
            all+=mp[d][i];
    group[d]=2;
    dfs(d+1,sum+all);
    all=0;
    for (int i=0;i!=d;i++)
        if (group[i]==2)
            all+=mp[d][i];
    group[d]=1;
    dfs(d+1,sum+all);
}

int main()
{
    scanf("%d",&N);
    for (int i=0;i<N;i++)
        for (int j=0;j<N;j++)
            scanf ("%d",&mp[i][j]);
    ans=-1;
    dfs(0,0);
    printf("%d\n",ans);
    return 0;
}
/*
3
0 50 30
50 0 40
30 40 0
*/