上一页 1 ··· 13 14 15 16 17 18 19 20 21 ··· 40 下一页
摘要: 题目链接: "http://poj.org/problem?id=2960" S Nim Time Limit: 2000MS Memory Limit: 65536K 问题描述 Arthur and his sister Caroll have been playing a game called 阅读全文
posted @ 2016-07-29 23:20 fenicnn 阅读(134) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://acm.hdu.edu.cn/showproblem.php?pid=5446" Unknown Treasure 问题描述 On the way to the next secret treasure hiding place, the mathematician di 阅读全文
posted @ 2016-07-29 23:13 fenicnn 阅读(178) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://codeforces.com/contest/161/problem/D" D. Distance in Tree time limit per test 3 secondsmemory limit per test 512 megabytes 问题描述 A tree i 阅读全文
posted @ 2016-07-29 22:27 fenicnn 阅读(122) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://codeforces.com/problemset/problem/242/E" E. XOR on Segment time limit per test 4 secondsmemory limit per test 256 megabytes 问题描述 You've 阅读全文
posted @ 2016-07-29 22:13 fenicnn 阅读(318) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://codeforces.com/problemset/problem/353/D?mobile=true" H. Queue time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 There 阅读全文
posted @ 2016-07-29 22:00 fenicnn 阅读(441) 评论(0) 推荐(0) 编辑
摘要: 0、目录 ac自动机、后缀数组、Trie树、kmp、回文串算法、后缀自动机、字符串哈希 1、ac自动机 include include include include using namespace std; const int maxn=1000+10; const int maxm=50+10; 阅读全文
posted @ 2016-07-29 00:21 fenicnn 阅读(140) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://codeforces.com/contest/103/problem/D" D. Time to Raid Cowavans time limit per test:4 secondsmemory limit per test:70 megabytes 问题描述 As y 阅读全文
posted @ 2016-07-28 23:00 fenicnn 阅读(246) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://codeforces.com/problemset/problem/268/E" E. Playlist time limit per test 1 secondmemory limit per test 256 megabytes 问题描述 Manao's friend 阅读全文
posted @ 2016-07-27 23:18 fenicnn 阅读(286) 评论(0) 推荐(0) 编辑
摘要: 题目链接: "http://codeforces.com/contest/435/problem/D" D. Special Grid time limit per test:4 secondsmemory limit per test:256 megabytes 问题描述 You are give 阅读全文
posted @ 2016-07-26 23:51 fenicnn 阅读(224) 评论(0) 推荐(0) 编辑
摘要: 0、目录 最短路、最小生成树、LCA (参考自白皮) 1、最短路 1.1、Floyd for(int i=1;i path; path.push_back(t); while(pre[s][t]!=s) { path.pb(pre[s][t]); t=pre[s][t]; } path.push_b 阅读全文
posted @ 2016-07-25 00:20 fenicnn 阅读(239) 评论(0) 推荐(0) 编辑
上一页 1 ··· 13 14 15 16 17 18 19 20 21 ··· 40 下一页