Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17349   Accepted: 7304 Description 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 currenc…

题目链接:http://poj.org/problem?id=2240. Floyd算法修改一下,我要最大路径(通过转汇率变到最大)改成max. #include <iostream> #include <string.h> #include <stdio.h> #include <algorithm> using namespace std; ][]; int n; int main() { ; while(scanf("%d",&am…

http://poj.org/problem?id=2240 题意:有些人会利用货币的不用汇率来进行套现,比如1美元换0.5英镑,而1英镑又可以换10法郎,而1法郎又可以换0.21的美元,那么经过货币的汇率转换后,它就可以获得1.05倍原来的美元. 现在给你N中货币,m种货币的汇率,求是否可以获利. 思路:首先这个是给你一些货币,那么我们可以把货币和数字建立一个映射.然后再用他给的汇率以及数字建立一个邻接矩阵.用一次floyd后对对角线的数字进行判断,如果大于1,那么说明可以获利,不然就不能获利…

是一道floyd变形的题目.题目让确定有几个人的位置是确定的,如果一个点有x个点能到达此点,从该点出发能到达y个点,若x+y=n-1,则该点的位置是确定的.用floyd算发出每两个点之间的距离,最后统计时,若dis[a][b]之间无路且dis[b][a]之间无路,则该点位置不能确定.最后用点个数减去不能确定点的个数即可.题目: Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4813   Accep…

链接:poj 2240 题意:首先给出N中货币,然后给出了这N种货币之间的兑换的兑换率. 如 USDollar 0.5 BritishPound 表示 :1 USDollar兑换成0.5 BritishPound. 问在这N种货币中是否存在货币经过若干次兑换后,兑换成原来的货币能够使货币量添加. 思路:本题事实上是Floyd的变形.将变换率作为构成图的路径的权值.只是构成的图是一个有向图. 最后将松弛操作变换为:if(dis[i][j]<dis[i][k]*dis[k][j]). #includ…

题目链接:http://poj.org/problem?id=3660 Description N ( ≤ N ≤ ) cows, conveniently numbered ..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique am…

链接 Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Eac…

Description An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D.…

题目: BIT has recently taken delivery of their new supercomputer, a 32 processor Apollo Odyssey distributed shared memory machine with a hierarchical communication subsystem. Valentine McKee's research advisor, Jack Swigert, has asked her to benchmark…

题意:奶牛拍一系列电影,n头牛拍m部电影,同一部电影种的搭档们距离为1,求最小距离? 思路:Floyd 图 最短路径 存图: 初始化图 for (int i = 0; i <= n; i++) { for (int j = 0; j <= n; j++) f[i][j] = INF; f[i][i] = 0; } 存图: for (int i = 0; i < k; i++) { scanf("%d", &a[i]); for (int j = 0; j &l…