POJ2531:Network Saboteur

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

30 50 3050 0 4030 40 0

Sample Output

90

这题的题意比较难看懂，将图中的n个点分为两个部分，求出一边所有点到另一边所有点的距离最大值是什么值

只需要枚举所有状况，因为n最大只有20，所以也不用担心超时

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int map[25][25],set[25];int main(){    int n,i,j,k;    while(~scanf("%d",&n))    {        for(i = 0; i<n; i++)            for(j = 0; j<n; j++)                scanf("%d",&map[i][j]);        memset(set,0,sizeof(set));        int maxn = 1<<(n-1);        int ans = 0;        for(i = 0; i<maxn; i++)//枚举2^n-1次就可以将所有情况枚举出来        {            set[0]++;            for(j = 0; j<n; j++)//通过0,1的值将n个点分到两个部分            {                if(set[j] == 2)                {                    set[j] = 0;                    set[j+1]++;                }                else break;            }            int cnt = 0;            for(j = 0; j<n; j++)//求出总和            {                if(!set[j])                    continue;                for(k = 0; k<n; k++)                    if(!set[k])                        cnt+=map[j][k];            }            ans = max(ans,cnt);        }        printf("%d\n",ans);    }    return 0;}

	
				
	
					   
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