Treap+set仿函数重定义.每当ac一道题目时,相当于对总时间减去一个大数. /* 5096 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algor…

title: [线性代数]3-3:秩(Rank) categories: Mathematic Linear Algebra keywords: Rank Row Reduced form Pivot Columns Free Columns Special Solutions toc: true date: 2017-09-25 15:20:38 Abstract: 本文将介绍线性代数中最最最重要的概念之一,秩(Rank) Keywords: Rank,Row Reduced form,Piv…

Friends and Enemies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 291    Accepted Submission(s): 160 Problem Description On an isolated island, lived some dwarves. A king (not a dwarf) ruled t…

Function Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 976    Accepted Submission(s): 375 Problem Description The shorter, the simpler. With this problem, you should be convinced of this tru…

Football Games Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 802    Accepted Submission(s): 309 Problem Description A mysterious country will hold a football world championships---Abnormal Cup…

Learning to Rank入门小结 + 漫谈 Learning to Rank入门小结 Table of Contents 1 前言 2 LTR流程 3 训练数据的获取4 特征抽取 3.1 人工标注 3.2 搜索日志 3.3 公共数据集 5 模型训练 5.1 训练方法 5.1.1 Pointwise 5.1.2 Pairwise 5.1.3 Listwise 6 效果评估7 参考 6.1 NDCG(Normalized Discounted Cumulative Gain) 6.1.1 定…

并查集+拓扑排序.使用并查集解决a = b的情况. #include <iostream> #include <cstdio> #include <cstring> #include <queue> using namespace std; #define MAXN 10005 typedef struct ArcNode { int adjvex; ArcNode *next; } ArcNode; typedef struct VNode { int c…

STL的使用. /* 5131 */ #include <iostream> #include <map> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> using namespace std; #define MAXN 205 typedef struct { string s;…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

挺有意思的一道题,思路肯定是将图转化为Trie树,这样可以求得字典序.然后,按照trie的层次求解.一直wa的原因在于将树转化为线性数据结构时要从原树遍历,从trie遍历就会wa.不同结点可能映射为trie上的同一结点,如1->2 (a) 1->3(a) 2->4(b), 这是trie的结构是RT->a->b.然而,从结点3不能找到权重为b的路径.用RMQ求满足边界的rank最大值,通过sa找到该最大值对应的trie上的根.从而求解. /* 4601 */ #include…