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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learn, build, evaluate
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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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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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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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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1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1072 1072. Routing Time limit: 1.0 second Memory limit: 64 MB There is a TCP/IP net of several computers. It means that: Each computer has one or…
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1. 题目 http://poj.org/problem?id=3414 Pots Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12107 Accepted: 5110 Special Judge Description You are given two pots, having the volume of A and B…
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1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1080 1080. Map Coloring Time limit: 1.0 second Memory limit: 64 MB We consider a geographical map with N countries numbered from 1 to N (0 < N…
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1. 题目 http://acm.timus.ru/problem.aspx?space=1&num=1033 1033. Labyrinth Time limit: 1.0 second Memory limit: 64 MB Administration of the labyrinth has decided to start a new season with new wallpa…
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