Time Limit: 2000MS Memory Limit: 32768K Total Submissions: 20754 Accepted: 10872 Description A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount
EK算法: int fir[maxn]; int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm]; int e_max; int p[maxn],q[maxn],d[maxn]; void add_edge(int _u,int _v,int _w) { int e; e=e_max++; u[e]=_u;v[e]=_v;cap[e]=_w; nex[e]=fir[u[e]];fir[u[e]]=e; e=e_max++; u[e]=_v;v[e]=
Drainage Ditches Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 70588 Accepted: 27436 Description Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by
题意: 第一行输入N M C ,表示从1到N有M条无向边,现在要从1走到N 走C次完全不同的路径,求最长边的最小值.下面M行是从a点到b点的距离. 建图: 题上说从两点之间可以有多条边,问的是从1~N的C种走法,所走路径上的最大边最小可以是多少,所以我们用结构体来储存点的距离,用二分搜索中的mid来假设成最大边的值,那么其他边都要小于等于它,然后根据mid建图,两点之间的边数作为容量网络的边权,也就是两点间的容量,(Dinic跑一遍)求出一个最大流,看它与C的关系,如果C大于它,说明mid太小,
题目:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Flow Problem Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 5111 Accepted Submission(s): 2385 Problem Description Network flow is a well-kn
对偶上升法 增广拉格朗日乘子法 ADMM 交替方向乘子法(Alternating Direction Method of Multipliers,ADMM)是一种解决可分解凸优化问题的简单方法,尤其在解决大规模问题上卓有成效,利用ADMM算法可以将原问题的目标函数等价的分解成若干个可求解的子问题,然后并行求解每一个子问题,最后协调子问题的解得到原问题的全局解,适用于大规模分布式优化问题. Lasso的ADMM求解算法