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…

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…

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…

周三的算法课,主要讲了随机化算法,介绍了拉斯维加斯算法,简单的理解了为什么要用随机化算法,随机化算法有什么好处. 在处理8皇后问题的时候,穷举法是最费时的,回朔比穷举好点,而当数据量比较大的时候,如1000皇后问题,穷举的化有1000的1000次方,肯定超时,用随机化算法的思路,先随机的在棋盘上放一部分皇后,再来设计算法,会大大节省时间,使算法性能更加优良. 本篇介绍令一种随机化算法,舍伍德(Sherwood)算法. 题目: Matrix Multiplication Time Limit: 2…

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…

Description You are given three n × n matrices A, B and C. Does the equation A × B = C hold true? Input The first line of input contains a positive integer n (n ≤ 500) followed by the the three matrices A, B and C respectively. Each matrix's descript…

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…

传送门 根据国家集训队2014论文集中胡泽聪的随机化算法可以通过这道题. 对于每个数,它有12" role="presentation" style="position: relative;">1212的概率在最后的答案序列中,这样我们每次随机出序列中的一个数,然后看它的因子有没有符合条件的更新答案就行了. 代码: #include<bits/stdc++.h> #define ll long long #define N 1000005…

题目大意:原题链接 给定三个n*n的矩阵A,B,C,验证A*B=C是否成立. 所有解法中因为只测试一组数据,因此没有使用memset清零 Hint中给的傻乎乎的TLE版本: #include<cstdio> #include<cstring> ][]; ][],C[][]; ][]) { ;i<=n;i++){ ;j<=n;j++) scanf("%d",&m[i][j]); } } int main() { scanf("%d&q…