ACM算法讲解宣讲培训.ppt

依次类推，经过n次比较和修正，在第（n-1）步，将求得vi到vj 的且中间顶点号不大于n-1的最短路径，这必是从vi到vj的最短路径。 图g中所有顶点偶对vi、vj间的最短路径长度对应一个n阶方阵D。在上述n+1步中，D的值不断变化，对应一个n阶方阵序列。 for(k=0;kn;++k){ for(i=0;in;++i){ for(j=0;jn;++j){ if(cost[i][j]cost[i][k]+cost[k][j]){ cost[i][j]=cost[i][k]+cost[k][j] } } } } Floyd-warshell 算法时间复杂度为O(n^3) 原程序实际上是这个DP的一个精巧的实现，省略了最后一维数组 练习题目 Toj 2870 The K-th City 单纯的Dijkstra Toj 2878 Internet is Faulty 先Floyed再Dijkstra, 而且路径权值是用乘法计算,可以直接把算法中的+变成*来解决,或者取对数,变成+计算亦可. Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pound, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollar. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent. Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not. Input Specification The input file will contain one or more test cases. On the first line of each test case there is an integer n (1=n=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Exchanges which do not appear in the table are impossible.Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n. Output Specification For each test case, print one line telling whether arbitrage is possible or not in the format Case case: Yes respectively Case case: No Arbitrage（套利） 解题报告 Sample Input 3 USDollar BritishPound FrenchFranc 3 USD