POJ 2531 Network Saboteur（dfs）

题目代号：POJ 2531

题目链接：http://poj.org/problem?id=2531

Language: Default

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13290		Accepted: 6428

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

题目大意：这个题就是题目有点难以理解，相当于有n台服务器，现在要将n台服务器分配成2个区域，不同区域之间的任意两台服务器会通过两个区域之间的节点交换数据，现在要你求，节点通过的最大数据量是多少，第i行第j列的数值相当于第i台服务器与第j台服务器交换的数据量。

解题思路：dfs遍历一遍，每遍历一次获取最大值，最后输出就行了。

AC代码：

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;

int a[25][25],ans,n;
int b[25];

void dfs(int num,int data)
{
    int s=data;
    b[num]=1;
    for(int i=1;i<=n;i++)
    {
        if(b[i])data-=a[num][i];
        else data+=a[num][i];
    }
    ans=max(ans,data);
    for(int i=num+1;i<=n;i++)
    {
        if(data>s)
        {
            dfs(i,data);
            b[i]=0;
        }
    }
    return;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                scanf("%d",&a[i][j]);
        ans=0;
        mem(b,0);
        dfs(1,0);
        cout<<ans<<endl;
    }
    return 0;
}