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POJ 3617 Best Cow Line (贪心)
Best Cow Line Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11230   Accepted: 3329 Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the an...
分类:其他好文   时间:2015-01-22 13:24:57    阅读次数:157
POJ3278Catch That Cow(线性模型)(BFS)
Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 49988   Accepted: 15679 Description Farmer John has been informed of the location of a fugit...
分类:其他好文   时间:2015-01-21 22:38:55    阅读次数:259
raw,cow,qcow,qcow2镜像的比较
镜像格式比较
分类:其他好文   时间:2015-01-19 17:10:11    阅读次数:206
POJ 3281-Dining(网络流_最大流_ISAP算法+拆点)
Dining Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9938   Accepted: 4567 Description Cows are such finicky eaters. Each cow has a preference for certai...
分类:编程语言   时间:2015-01-19 09:22:19    阅读次数:208
HLJU14级寒假集训之最短路专场
下面是今天的题解链接,感觉写的还是很详细的,有什么问题可以找我,大家一起探讨~~~ A - MPI Maelstrom: 题解链接:MPI Maelstrom B - Cow Hurdles: 题解链接:Cow Hurdles C - Wormholes: 题解链接:Wormholes D - Frogger: 题解链接:Frogger...
分类:其他好文   时间:2015-01-18 21:08:40    阅读次数:145
(简单) POJ 1278 Catch That Cow,回溯。
Description Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100...
分类:其他好文   时间:2015-01-16 12:57:44    阅读次数:118
(简单) POJ 3279 Fliptile,集合枚举。
Description Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for co...
分类:其他好文   时间:2015-01-16 12:50:45    阅读次数:183
POJ 3615 Cow Hurdles (Floyd算法)
Cow Hurdles Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6142   Accepted: 2752 Description Farmer John wants the cows to prepare for the county jumpin...
分类:编程语言   时间:2015-01-15 00:30:35    阅读次数:247
C++算法之 求二叉树两个节点的最低公共节点
方法1:递归方法: (1)如果两个节点分别在根节点的左子树和右子树,则返回根节点 (2)如果两个节点都在左子树,则递归处理左子树;如果两个节点都在右子树,则递归处理右子树 bool FindNode(BTree* pRoot, BTree* pNode) { if (pRoot == NULL || pNode == NULL) { return false; } ...
分类:编程语言   时间:2015-01-13 10:36:26    阅读次数:231
C++算法之 判断是否为完全二叉树
判断完全二叉树: 完全二叉树,除最后一层外,每一层上的节点树都达到了最大值;在最后一层上只缺少右边的若干节点! 算法思路: 按层次(从上到下,从左到右)遍历二叉树,当遇到一个节点的左子树为空时,则该节点右子树必须为空,且后面遍历的节点左 右子树都必须为空,否则不是完全二叉树。 代码: bool IsCompleteBTree(BTree* pRoot) { if (pRoot =...
分类:编程语言   时间:2015-01-12 13:06:28    阅读次数:475
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