POJ 2531-Network Saboteur(dfs+剪枝)

浏览数：35 / 时间：2015年06月09日

Network Saboteur

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 9544		Accepted: 4542

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

Sample Output

题意：给出n个点(0-n-1)，接下来n行，每行有n个数，第i行代表第i-1个数距离n个点的距离，问将这些点分为两部分，两个点集之间的权值最大值为多少。例如样例中讲0,1,2分为两部分（1，3）和（2）两部分，则最长距离为map[1][2]+map[3][2]=90。

思路：怎么想都没看出这题和dfs有什么关系，知道看了巨巨的题解，恍然大悟，涨姿势了。首先我们将所有的店标记为0(即所有点放在一个集合里)，然后取出一个site标记为1(即将该点放在另一个集合里)，这是对于和site在一个集合里的点，我们减去他们两个之间的权值，对于不在一个集合里的点，我们加上他们之间的权值。最后的结果为最大值。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>

using namespace std;
const int inf=0x3f3f3f3f;
int map[40][40];
int g[40];
int n,res;
void dfs(int site,int sum)
{
    int i;
    g[site]=1;//取出点
    int num=sum;
    for(i=0;i<n;i++){
        if(g[i]==1)//与site在一个集合里的点
            num-=map[site][i];//减去他们之间的权值
        else//否则和site不在一个集合里
            num+=map[site][i];//加上他们之间的权值
    }
    res=max(res,num);//依次循环结束找到的最大值
    for(i=site+1;i<n;i++){//然后遍历剩下的点
        if(num>sum){//小剪枝，如果加入了点site后两个集合之间的权值小了，则就不需要遍历这个点。
            dfs(i,num);
            g[i]=0;
        }
    }

}
int main()
{
    int i,j;
    while(~scanf("%d",&n)){
        memset(map,0,sizeof(map));
        memset(g,0,sizeof(g));
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            scanf("%d",&map[i][j]);
            res=-inf;
            dfs(0,0);
            printf("%d\n",res);
    }
    return 0;
}

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POJ 2531-Network Saboteur(dfs+剪枝)