poj2531:http://poj.org/problem?id=2531 题意:给你一个图,图中点之间会有边权,现在问题是把图分成两部分,使得两部分之间边权之和最大.题解:随机算法 #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; int n; ][]; ]; ;//时间限制是2000 int main(){ sca…

Network Saboteur DescriptionA 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…

Network Saboteur Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11122   Accepted: 5372 Description A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully d…

Network Saboteur Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10147 Accepted: 4849 Description A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully divid…

C - Network Saboteur Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Description A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and carefully divide…

Network Saboteur POJ2531 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10351   Accepted: 4968 Description A university network is composed of N computers. System administrators gathered information on the traffic between nodes, and car…

                            Network Assistant (网络助手)使用说明 本软件是一款帮助同学方便Ip更改,小蝴蝶断线重连,一键wifi,定时关机的软件.它集成了这四项功能,可以帮助一些电脑小白. 尤其是刚刚入学的大一新生,以及非信息学院的学生使用. 这个本软件主界面: 校园认证功能: 只需输入用户名,密码,点击认证,便可以连上校园内网. 网络地址转换器: 可以在“网卡信息”中填写网卡信息,点击“修改生效”,将更改网卡,点击“保存为新方案”,将保存为一个方…

题目: 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 sc…

题目大意:原题链接 给定n个节点,任意两个节点之间有权值,把这n个节点分成A,B两个集合,使得A集合中的每一节点与B集合中的每一节点两两结合(即有|A|*|B|种结合方式)权值之和最大. 标记:A集合:true  B集合:false 解法一:dfs+剪枝 #include<iostream> #include<cstring> using namespace std; int n,ans; ]; ][]; void dfs(int i,int cursum) { in[i]=tru…

个人心得:对于深搜的使用还是不到位,对于递归的含义还是不太清楚!本来想着用深搜构成一个排列,然后从一到n分割成俩个数组,然后后面发现根本实现不了,思路太混乱.后来借鉴了网上的思想,发现用数组来标志,当值等于一时就属于A数组,等于0时属于B数组,这样就可以构成递归,即下一个数只有在A数组和不在A数组,发现这个思想真的挺好的. DFS心得:若从第二步开始,动作变得程序化,像地图,只有上下左右,这个是只有为0不为0是就可以用深搜递归思想解决. A university network is compo…