POJ2531Network Saboteur(DFS+剪枝）

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 10391		Accepted: 4990

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
题意：将n个数分成两个集合，求一个集合中所有数到另一个集合所有数的最大和

#include <iostream>
#include <cstring>
#include <string.h>
#include <algorithm>
#include <cstdio>

using namespace std;
const int MAX = 22;
int a[MAX][MAX],v[MAX],b[MAX];
int n,cnt,sum;
void dfs(int x,int sum)
{
    v[x] = 1;
    for(int i = 1; i <= n; i++)
    {
        if(v[i] == 0)
            sum += a[x][i];
        else
            sum -= a[x][i];
    }
    cnt = max(cnt,sum);
    for(int i = x + 1; i <= n; i++) // 这个注意；其实可以这么想，以前求的是路径所以颠倒顺序也是一种情况，这是集合，所以颠倒顺序和原来是一种情况，所以没必要从0开始循环，0，1和1，0是一种情况
    {
        dfs(i,sum);
        v[i] = 0;
    }
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
            {
                scanf("%d", &a[i][j]);
            }
        }
        memset(v,0,sizeof(v));
        cnt = 0;
        dfs(1,0);
        printf("%d
",cnt);
    }
    return 0;
}