Jack Ge - 博客园

2013年5月2日

摘要： Buy the TicketTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3286Accepted Submission(s): 1366Problem DescriptionThe "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go t 阅读全文

posted @ 2013-05-02 22:37 Jack Ge 阅读(217) 评论(0) 推荐(0) 编辑

POJ 3160 Father Christmas flymouse （tarjan+spfa）

摘要： Father Christmas flymouseTime Limit:1000MSMemory Limit:131072KTotal Submissions:1999Accepted:674DescriptionAfter retirement as contestant from WHU ACM Team, flymouse volunteered to do the odds and ends such as cleaning out the computer lab for training as extension of his contribution to the team. W 阅读全文

posted @ 2013-05-02 07:45 Jack Ge 阅读(321) 评论(0) 推荐(0) 编辑

POJ 2762 Going from u to v or from v to u?

摘要： Going from u to v or from v to u?Time Limit:2000MSMemory Limit:65536KTotal Submissions:12260Accepted:3139DescriptionIn order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, 阅读全文

posted @ 2013-05-02 07:10 Jack Ge 阅读(339) 评论(0) 推荐(0) 编辑

2013年5月1日

POJ 3255 Roadblocks (第K短路 & A*)

摘要： RoadblocksTime Limit:2000MSMemory Limit:65536KTotal Submissions:5256Accepted:2013DescriptionBessie has moved to a small farm and sometimes enjoys returning to visit one of her best friends. She does not want to get to her old home too quickly, because she likes the scenery along the way. She has dec 阅读全文

posted @ 2013-05-01 23:11 Jack Ge 阅读(374) 评论(0) 推荐(0) 编辑

无向图割点模板 (转载)

摘要： 1.定义:割点：某个点是割点当且仅当删除该点和与该点相关联的边后图变得不连通。桥（割边）：某条边是割边当且仅当删除该边后图变的不连通。双连通分支：图G的所有子图G'中，如果G'是连通的，则称G'是双连通子图。如果G'不是任何一个连通子图的真子图，那么图G'是双连通分支。特别的，点双连通分支又叫做块。2.求割点，桥对原图进行深度优先搜索，会生成一颗深度优先搜索生成树。定义dfs[u]为u在深度优先搜索生成树中被遍历到的序号，low[u]为u或者他的子树中可以通过非父子边追溯到的最早结点。那么一个顶点是割点，满足下列条件之一：1）.u是树根，u有两个或两个以阅读全文

posted @ 2013-05-01 22:41 Jack Ge 阅读(450) 评论(0) 推荐(0) 编辑

POJ 1144 Network （割点，两种模板)

摘要： NetworkTime Limit:1000MSMemory Limit:10000KTotal Submissions:7510Accepted:3525DescriptionA Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional 阅读全文

posted @ 2013-05-01 22:36 Jack Ge 阅读(2725) 评论(0) 推荐(0) 编辑

POJ 3177 Redundant Paths (双连通)

摘要： Redundant PathsTime Limit:1000MSMemory Limit:65536KTotal Submissions:6982Accepted:3042DescriptionIn order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bessie and the rest of the herd are forced to cross near the Tree of Rotten Apples. Th 阅读全文

posted @ 2013-05-01 22:21 Jack Ge 阅读(472) 评论(0) 推荐(0) 编辑

POJ 2117 Electricity (割点)

摘要： ElectricityTime Limit:5000MSMemory Limit:65536KTotal Submissions:2515Accepted:839DescriptionBlackouts and Dark Nights (also known as ACM++) is a company that provides electricity. The company owns several power plants, each of them supplying a small area that surrounds it. This organization brings a 阅读全文

posted @ 2013-05-01 22:06 Jack Ge 阅读(592) 评论(0) 推荐(0) 编辑

HDU 4337 King Arthur's Knights

摘要： King Arthur's KnightsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1358Accepted Submission(s): 583Special JudgeProblem DescriptionI am the bone of my sword. Steel is my body, and the fire is my blood.- from Fate / Stay NightYou must have know 阅读全文

posted @ 2013-05-01 21:48 Jack Ge 阅读(347) 评论(0) 推荐(0) 编辑

Problem F: [USACO 3.1.6]邮票

摘要： Problem F: [USACO 3.1.6]邮票http://www.acmore.net/problem.php?id=1084Time Limit:1 SecMemory Limit:64 MBSubmit:35Solved:13SubmitStatusWeb BoardDescription已知一个 N 枚邮票的面值集合（如，{1 分，3 分}）和一个上限 K —— 表示信封上能够贴 K 张邮票。计算从 1 到 M 的最大连续可贴出的邮资。例如，假设有 1 分和 3 分的邮票；你最多可以贴 5 张邮票。很容易贴出 1 到 5 分的邮资（用 1 分邮票贴就行了），接下来的邮资也不难：阅读全文

posted @ 2013-05-01 19:08 Jack Ge 阅读(202) 评论(0) 推荐(0) 编辑

Jack Ge for ACM

感谢对手，你们令我不断进取、努力，，感谢朋友，你们让我感受到世界的温暖、快乐

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