zoukankan      html  css  js  c++  java
  • POJ 2531-Network Saboteur(DFS)

    Network Saboteur
    Time Limit: 2000MS   Memory Limit: 65536K
    Total Submissions: 9435   Accepted: 4458

    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n个点和他们之间的权值Cij。如今要将n个点分为两部分,求∑Cij (i∈A,j∈B)最大
    思路:dfs暴搜
    算是没剪枝吧。。跑了219MS sad
    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <string>
    #include <cctype>
    #include <vector>
    #include <cstdio>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <map>
    #include <set>
    #define ll long long
    #define maxn 360
    #define pp pair<int,int>
    #define INF 0x3f3f3f3f
    #define max(x,y) ( ((x) > (y)) ?

    (x) : (y) ) #define min(x,y) ( ((x) > (y)) ? (y) : (x) ) using namespace std; int ma[23][23],n,ans; bool vis[23]; void dfs(int x,int num,int sum) { if(num>n/2+1)return ; int t=0; ans=max(ans,sum); for(int i=x+1;i<=n;i++) { if(!vis[i]) { vis[i]=1; int tem=sum; for(int j=1;j<=n;j++) if(vis[j]) tem-=ma[i][j]; else tem+=ma[i][j]; dfs(i,num+1,tem); vis[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",&ma[i][j]); memset(vis,0,sizeof(vis)); ans=-INF; dfs(0,0,0); printf("%d ",ans); } return 0; }

    版权声明:本文博客原创文章。博客,未经同意,不得转载。

  • 相关阅读:
    岛田庄司《占星术杀人魔法》读后感
    OutputCache祥解
    ZOJ Monthly, June 2014 月赛BCDEFGH题题解
    接口和抽象类有什么差别
    C语言指针的初始化和赋值
    深入探讨this指针
    郁 繁体为“鬰” 古同 “鬱”
    socketpair的使用
    Android的FrameLayout使用要注意的问题
    下确界和上确界
  • 原文地址:https://www.cnblogs.com/blfshiye/p/4666438.html
Copyright © 2011-2022 走看看