昨天学到了一个新的算法tarjan算法,感觉最近都没有怎么学习了。。。(最近有个感悟啊,就是学习一定的通过实践来进步的。 现在才明白为什么高中的时候老师
https://www.u72.net/daima/820s.html - 2024-09-12 04:13:46 - 代码库题目链接:http://61.187.179.132/JudgeOnline/problem.php?id=2438题意:一位冷血的杀手潜入某村庄,并假装成 平民。警察希望能在 N 个人里面,查出谁是杀手
https://www.u72.net/daima/bua3.html - 2024-07-09 00:02:37 - 代码库题意:给定一个有向图,问有多少个点由任意顶点出发都能达到。分析:首先,在一个有向无环图中,能被所有点达到点,出度一定是0。先求出所有的强连通分支,然后把每
https://www.u72.net/daima/cmdk.html - 2024-07-11 14:27:25 - 代码库迷宫城堡Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6655 Accepted Submission
https://www.u72.net/daima/rzrf.html - 2024-07-11 17:13:10 - 代码库题目地址:ZOJ 2588因为数组开小了而TLE了。。这题就是一个求无向连通图最小割边。只要判断dfn[u]是否<low[v],因为low指的当前所能回到的祖先的最小标号,
https://www.u72.net/daima/15eu.html - 2024-07-19 11:49:25 - 代码库题目地址:POJ 1236这个题的大意是求最少往多少点发送消息可以使任意一个点都能收到消息和最少增加多少条边可以使图为连通图。对于第一个问题,可以求入度
https://www.u72.net/daima/15n9.html - 2024-07-19 10:59:04 - 代码库Redundant PathsTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13712 Accepted: 5821DescriptionIn order to get fr
https://www.u72.net/daima/wm3x.html - 2024-08-26 12:00:34 - 代码库Intelligence SystemTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1386 Accepted
https://www.u72.net/daima/wux3.html - 2024-07-16 03:30:05 - 代码库Summer HolidayTime Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1626 Accepted Sub
https://www.u72.net/daima/wu59.html - 2024-07-16 03:40:55 - 代码库POJ 2186 Popular Cows 链接:http://poj.org/problem?id=2186题意:每头奶牛都梦想着成为牧群中最受奶牛仰慕的奶牛。在牧群中,有N 头奶牛,1≤N≤10,000,给
https://www.u72.net/daima/3xvk.html - 2024-07-21 08:16:57 - 代码库POJ 1236 Network of Schools链接:http://poj.org/problem?id=1236题意:有一些学校连接到一个计算机网络。这些学校之间达成了一个协议:每个学校维护着一
https://www.u72.net/daima/30fs.html - 2024-07-21 08:59:37 - 代码库【题目链接】 http://poj.org/problem?id=2186 【题目大意】 给出一张有向图,问能被所有点到达的点的数量 【题解】 我们发现能成为答案的,只有拓
https://www.u72.net/daima/8axw.html - 2024-09-11 02:11:30 - 代码库POJ 1236 Network of Schools题目链接题意:题意本质上就是,给定一个有向图,问两个问题1、从哪几个顶点出发,能走全所有点2、最少连几条边,使得图强连通思路:#
https://www.u72.net/daima/eh08.html - 2024-07-28 05:10:00 - 代码库NetworkTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 6837 Accepted: 2435DescriptionA network administrator manages a large netwo
https://www.u72.net/daima/83m8.html - 2024-07-26 17:08:30 - 代码库传送门:Cactus判断给定的有向图是否满足 1.强连通 2 每一条边属于且仅属于一个环?YES:NO存在有两种情况(yy一下)1.他的子节点在栈中2.他的子节点的最早的时
https://www.u72.net/daima/nbwv2.html - 2024-08-06 05:41:50 - 代码库看了别人博客 http://blog.csdn.net/jokes000/article/details/7538994 #include <cstdio>#include <cmath>#include <algorithm>#include <iostream>
https://www.u72.net/daima/nzf1v.html - 2024-08-01 14:40:20 - 代码库传送门DescriptionA number of schools are connected to a computer network. Agreements have been developed among those schools: each school ma
https://www.u72.net/daima/z793.html - 2024-08-12 22:43:25 - 代码库Redundant PathsDescriptionIn order to get from one of the F (1 <= F <= 5,000) grazing fields (which are numbered 1..F) to another field, Bes
https://www.u72.net/daima/2ufe.html - 2024-09-01 07:58:30 - 代码库Byteotia城市有n个 towns m条双向roads. 每条 road 连接 两个不同的 towns ,没有重复的road. 你要把其中一些road变成单向边使得:每个town都有且只有一
https://www.u72.net/daima/7bdk.html - 2024-09-09 17:45:51 - 代码库先缩点,对于缩完点后的DAG,可以直接在每个scc dfs一次就可以求出终点是这个scc的点的点对个数。 # include <cstdio># include <cstring># include
https://www.u72.net/daima/87u8.html - 2024-09-12 11:33:44 - 代码库