随笔分类 - ACM-图论-最短路
摘要:学习链式前向星:http://blog.csdn.net/acdreamers/article/details/16902023 HDU 1874 留个链式前向星+Dijkstra模板。
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摘要:HDU 5573 Binary Tree 获得姿势:要构造数,想到用二进制构造。 一直向最左路构造,最左路的和是(2^k)-1,加入给定n,那么被减去的数字是d = (2^k)-1-n. 当我要减去一个数字时,从上往下走,对于总和相当于减去二倍的数字,所以要d/2.但前提是保证d为偶数。即为奇数时d
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摘要:In Action Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5220 Accepted Submission(s): 1745 Probl
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摘要:XYZZY Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4421 Accepted Submission(s): 1252 Problem D
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摘要:一个人的旅行 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29572 Accepted Submission(s): 10160 Proble
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摘要:畅通工程续 Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 42384 Accepted Submission(s): 15689 Problem
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摘要:最短路径问题 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 20846 Accepted Submission(s): 6191 Problem
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摘要:最短路 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 50686 Accepted Submission(s): 22280 Problem D
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摘要:最近开始学习最短路算法。 单源最短路问题1(Bellman-Ford算法) 不存在负圈时, 最短路不会经过同一顶点两次(也就是说最多通过v-1条边) while(1)最多执行v-1次,时间复杂度o(v*e). 如果存在从s可达的负圈,那么在第v次循环中也会更新d的值,所以可以用来检查负圈。
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