和POJ1860差不多,就是用bellmanford判读是否存在正环,注意的是同种货币之间也可以交换,就是说:A货币换A货币汇率是2的情况也是存在的. #include<stdio.h> #include<string.h> #include<cstring> #include<string> #include<math.h> #include<queue> #include<algorithm> #include<…

Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions:27167   Accepted: 11440 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…

题目传送门 /* 最短路:Floyd模板题 只要把+改为*就ok了,热闹后判断d[i][i]是否大于1 文件输入的ONLINE_JUDGE少写了个_,WA了N遍:) */ #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #include <string> #include <map> #include <cmath>…

POJ 2240 Arbitrage / ZOJ 1092 Arbitrage / HDU 1217 Arbitrage / SPOJ Arbitrage(图论,环) 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 currency. For exa…

Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21300   Accepted: 9079 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…

链接: http://poj.org/problem?id=1860 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/A Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16244   Accepted: 5656 Description Several currency exchange point…

链接: http://poj.org/problem?id=2240 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=22010#problem/F Arbitrage Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13067   Accepted: 5493 Description Arbitrage is the use of discrepancies i…

题目:http://poj.org/problem?id=2240 题意:给定n个货币名称,给m个货币之间的汇率,求会不会增加 和1860差不多,求有没有正环 刚开始没对,不知道为什么用 double往结构体里传值的时候 会去掉小数点后的 数 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<qu…

题目链接:http://poj.org/problem?id=2240 题目就是要通过还钱涨自己的本钱最后还能换回到自己原来的钱种. 就是判一下有没有负环那么就直接用bellman_ford来判断有没有负环 #include <iostream> #include <cstring> #include <string> using namespace std; int n , m , s , a , b , counts; double rx , ry , cx , c…

d[i]代表从起点出发可以获得最多的钱数,松弛是d[v]=r*d[u],求最长路,看有没有正环 然后这题输入有毒,千万别用cin 因为是大输入,组数比较多,然后找字符串用strcmp就好,千万不要用map 这题刚开始我T了(用的map),还以为组数很多卡spfa呢,然后我上网看了看都是floyd的,然后我用floyd写了一发,891ms过了 然后我感觉spfa的复杂度也不是很大,就是看有没有正环,所以我觉得可能是map+cin的锅,然后改了一发,用的spfa,47ms过 真是,算了,实质是本蒟蒻…