URAL 1980. Road to Investor
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1980 1980. Road to Investor Time limit: 1.0 second Memory limit: 64 MB ‘How could I forget it? In T hours we must show the alpha-version of our g…
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URAL 1753. Bookshelf
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1753 1753. Bookshelf Time limit: 0.5 second Memory limit: 64 MB A bookshelf has length L and borders of height h. A book of height H is standing …
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POJ 3273. Monthly Expense
1. 题目 http://poj.org/problem?id=3273 Monthly Expense Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 18048 Accepted: 7226 Description Farmer John is an astounding accounting wizard and has …
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POJ 2456. Aggressive cows
1. 题目 http://poj.org/problem?id=2456 Aggressive cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8456 Accepted: 4224 Description Farmer John has built a new long barn, with N (2 <= N…
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POJ 1064. Cable master
1. 题目 http://poj.org/problem?id=1064 Cable master Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 30585 Accepted: 6482 Description Inhabitants of the Wonderland have decided to hold a regio…
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POJ 3259. Wormholes
1. 题目 http://poj.org/problem?id=3259 Wormholes Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 36967 Accepted: 13562 Description While exploring his many farms, Farmer John has discovered a…
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分治算法简介——归并排序
归并排序应用分治算法，具有O(nlogn)的时间复杂度，其工作流程可以概述为： 将长度为n的无序序列分割成n个子序列，每个序列包含1个元素； 重复将子序列相互合并称为有序序列，直到只剩下1个序列，即为排序后的序列。 一个例程如下所示： void merge(int array[], int tmpArray[], int lPos, int rPos, int rEnd) { int i,…
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Bellman-Ford算法简介
Bellman-Ford算法用于计算有向图中的单源最短路径。对于有V个顶点和E条边的图，Bellman-Ford的时间复杂度为O(V·E)，效率低于Dijkstra算法。Bellman-Ford可以用于有负权边的情况，且能判断负权环，十分灵活。 Wiki上给出的Bellman-Ford算法的伪代码如下所示： function BellmanFord(list vertices, list …
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URAL 1008. Image Encoding
1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1008 1008. Image Encoding Time limit: 2.0 second Memory limit: 64 MB There are several ways to encode an image. In this problem we will consider …
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