题目:http://acm.hdu.edu.cn/showproblem.php?pid=1217 /************************************************************************/ /* hdu Arbitrage floyd求最短路径 题目大意:floyd算法,求得两点之间最短距离,(u,v) = min( (u,v),(u,w)+(w,v) ); 此题,是求其能够赚钱,即最大生成树,且过程是相乘的,公式:( u, v) =…

UVA - 10048 Audiophobia Consider yourself lucky! Consider yourself lucky to be still breathing and having fun participating in this contest. But we apprehend that many of your descendants may not have this luxury. For, as you know, we are the dweller…

Frogger DescriptionFreddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimm…

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…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 Arbitrage Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4430    Accepted Submission(s): 2013 Problem Description Arbitrage is the use of discr…

http://acm.hdu.edu.cn/showproblem.php?pid=1596 find the safest road Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6911    Accepted Submission(s): 2450 Problem Description XX星球有很多城市,每个城市之间有一条或…

Arbitrage http://acm.hdu.edu.cn/showproblem.php?pid=1217 Problem 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 example, suppose that…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意:问你是否可以通过转换货币从中获利 如下面这组样例: USDollar 0.5 BritishPound BritishPound 10.0 FrenchFranc FrenchFranc 0.21 USDollar 可以通过US->Br->French->US这样转换,把1美元变成1*0.5*10*0.21=1.05美元赚取%5的利润. 解题思路:其实就相当于bellman-…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4034 Problem Description Everyone knows how to calculate the shortest path in a directed graph. In fact, the opposite problem is also easy. Given the length of shortest path between each pair of vertexe…

给出一些国家之间的汇率,看看能否从中发现某些肮脏的......朋友交易. 这是Floyd的应用,dp思想,每次都选取最大值,最后看看自己跟自己的.....交易是否大于一.... #include<iostream> #include<cstring> #include<queue> #include<cstdio> #include<map> using namespace std; #define exp 0.00000001 map<s…

题意:给几个国家,然后给这些国家之间的汇率.判断能否通过这些汇率差进行套利交易. Floyd的算法可以求出任意两点间的最短路径,最后比较本国与本国的汇率差,如果大于1,则可以.否则不可以. 有向图 一个点到另一点的花费为权值相乘 求乘积的最大值 从点i出发 再回到点i的花费如果大于1 就可以 Sample Input3USDollarBritishPoundFrenchFranc3USDollar 0.5 BritishPoundBritishPound 10.0 FrenchFrancFren…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1869 题意分析:比较简单的最短路算法,最后只需判断最远两点距离是否大于7即可. /*六度分离 Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4992 Accepted Submission(s): 2010 Problem Descriptio…

Rank Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1622    Accepted Submission(s): 625 Problem Description there are N ACMers in HDU team.ZJPCPC Sunny Cup 2007 is coming, and lcy want to selec…

find the safest road Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10450    Accepted Submission(s): 3694 Problem Description XX星球有很多城市,每个城市之间有一条或多条飞行通道,但是并不是所有的路都是很安全的,每一条路有一个安全系数s,s是在 0 和 1…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1217 题目大意:通过货币的转换,来判断是否获利,如果获利则输出Yes,否则输出No. 这里介绍一个STL中的map容器去处理数据,map<string,int>V,M; 现在我目前的理解是将字符串转换成数字,然后就是根据spfa的模板找最短路了..哇哈哈( ⊙o⊙ )哇 #include <iostream> #include <cstdio> #include <m…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 31458    Accepted Submission(s): 14128 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

变形课 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 18343    Accepted Submission(s): 6597 Problem Description 呃......变形课上Harry碰到了一点小麻烦,由于他并不像Hermione那样可以记住全部的咒语而任意的将一个棒球变成刺猬什么的,可是他发现了变形咒语的一个统一…

Arbitrage Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6360    Accepted Submission(s): 2939 Problem Description Arbitrage is the use of discrepancies in currency exchange rates to transform o…

Frogger Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2253 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone…

思路: 我们先求一遍floyd,将各点的最短距离求出,然后将点按si的升序排序.dp[i][k]表示第i个点在第j时间所获得的最大效益,那么 dp[i][k]=max(dp[ i ][ k ]  ,  dp[ j ][ k-p[ i ].c-dis[ i ][ j ] ]+p[ i ].s);     dis[i][j]为i与j的最短路径. #include<iostream> #include<cstdio> #include<cstring> #include<…

题意: 给一个带权无向图,和一些询问,每次询问两个点之间最大权的最小路径. 分析: 紫书上的题解是错误的,应该是把原算法中的加号变成max即可.但推理过程还是类似的,如果理解了Floyd算法的话,这个应该也很容易理解. #include <cstdio> #include <algorithm> using namespace std; + ; ; int d[maxn][maxn]; int main() { //freopen("in.txt", "…

Time Limit: 1000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1630    Accepted Submission(s): 664 Problem Description 杭州有N个景区,景区之间有一些双向的路来连接,现在8600想找一条旅游路线,这个路线从A点出发并且最后回到A点,假设经过的路线为V1,V2,....VK,V1,那么必须满足K>2,…

Advanced Fruits Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2358    Accepted Submission(s): 1201Special Judge Problem Description The company "21st Century Fruits" has specialized in cr…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30163    Accepted Submission(s): 13507 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

find the safest road Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7845    Accepted Submission(s): 2786 Problem Description XX星球有非常多城市.每一个城市之间有一条或多条飞行通道,可是并非全部的路都是非常安全的,每一条路有一个安全系数s,s是在 0 和…

Flody多源最短路 #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<map> #include<algorithm> #include<iostream> using namespace std; ; map<string,int>zh; string s,s1,s2; double jz[maxn][m…

XYZZY Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4421    Accepted Submission(s): 1252 Problem Description It has recently been discovered how to run open-source software on the Y-Crate gami…

一个人的旅行 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 29572    Accepted Submission(s): 10160 Problem Description 虽 然草儿是个路痴(就是在杭电待了一年多,居然还会在校园里迷路的人,汗~),但是草儿仍然很喜欢旅行,因为在旅途中 会遇见很多人(白马王子,^0^),很多事,还能…

Friends Time Limit: 2 Seconds      Memory Limit: 65536 KB Alice lives in the country where people like to make friends. The friendship is bidirectional and if any two person have no less thank friends in common, they will become friends in several da…

题目大意: 现在,已知起点和终点,请你计算出要从起点到终点,最短需要行走多少距离. 思路: floyd算法模板题,这是一个牺牲空间换取时间的算法,本质是动态规划. AC代码: #include <iostream> #include <cstdio> #include <string.h> using namespace std; +; const int INF = 0x3f3f3f3f; int n, m; int mp[MX][MX]; void floyd()…