POJ 2531 Network Saboteur (DFS)

Network Saboteur

 

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 12453 Accepted: 6009

 

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

给出一些点与它们之间所能消耗的流量,现在要把这些点分为两组,同一组的点之间无流量损耗,不同组的点之间需要消耗流量,求所能消耗的流量最大值。

 

思路

深度优先搜索

我们先假设所有点都在左边,现在把其中的点拿到右边,与它同组的点需要减去与它之间的流量,不同组的点加上这些流量,然后更新最大值。

如果一个点被拿到右边,改变之后的流量值小于它之前的流量值,则不要把其他点再移动到右边,该层递归结束。因为两个组是可以交换的,所做的变化满足对称性。

 

AC 代码

#include <iostream>
#include<stdio.h>
#include<stdlib.h>
#include<queue>
#include<string.h>
using namespace std;
int a[30][30];      //map
bool con[30];       //各点分组情况
int ans,n;

void dfs(int id,int tmp)    //将id移到右边,计算流量
{
    int old=tmp;
    con[id]=true;
    for(int i=0; i<n; i++)  //同组减,异组加
    {
        if(con[i])tmp-=a[i][id];
        else tmp+=a[i][id];
    }
    ans=max(ans,tmp);
    for(int i=id+1; i<n; i++)   //搜索后面的点
        if(!con[i]&&tmp>old)
            dfs(i,tmp);
    con[id]=false;
}

int main()
{
    while(~scanf("%d",&n))
    {
        ans=0;
        memset(con,false,sizeof(con));
        for(int i=0; i<n; i++)
            for(int j=0; j<n; j++)
                scanf("%d",&a[i][j]);
        for(int i=0; i<n; i++)
            dfs(i,0);
        printf("%d\n",ans);
    }
    return 0;
}