真是气skr人..没把d[]换成double...de了一上午的bug// 记得用G++提交啊 题目链接:http://poj.org/problem?id=1860 题意:告诉你n个点,m条路.起始点s,还有初始金额money.每条路表示从a->b的汇率和佣金以及b->a的汇率和佣金.你在该点所得是(本金-佣金)*汇率.问你这个人能不能赚钱. 题解:spfa套一下//.记得d[]换成double.具体的看看代码.QWQ. 代码: #include<iostream> #inclu…

题目传送门 /* 最短路(Bellman_Ford):求负环的思路,但是反过来用,即找正环 详细解释:http://blog.csdn.net/lyy289065406/article/details/6645778 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmat…

POJ 1860 Currency Exchange / ZOJ 1544 Currency Exchange (最短路径相关,spfa求环) Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations onl…

原题链接:http://poj.org/problem?id=1860 Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 23055   Accepted: 8328 Description Several currency exchange points are working in our city. Let us suppose that each point specializes…

三道题都是考察最短路算法的判环.其中1860和2240判断正环,3259判断负环. 难度都不大,可以使用Bellman-ford算法,或者SPFA算法.也有用弗洛伊德算法的,笔者还不会SF-_-…… 直接贴代码. 1860 Currency Exchange: #include <cstdio> #include <cstring> int N,M,S; double V; ; int first[maxn],vv[maxn*maxn],nxt[maxn*maxn]; double…

Description Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example: Georgia an…

Description 有两堆石子,数量任意,可以不同.游戏开始由两个人轮流取石子.游戏规定,每次有两种不同的取法,一是可以在任意的一堆中取走任意多的石子:二是可以在两堆中同时取走相同数量的石子.最后把石子全部取完者为胜者.现在给出初始的两堆石子的数目,如果轮到你先取,假设双方都采取最好的策略,问最后你是胜者还是败者. Input 输入包含若干行,表示若干种石子的初始情况,其中每一行包含两个非负整数a和b,表示两堆石子的数目,a和b都不大于1,000,000,000. Output 输出对应也有…

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…

Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the same pair o…

Currency Exchange Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1860 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two par…

Currency Exchange Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 1860 Appoint description:  System Crawler  (2015-05-14) Description Several currency exchange points are working in our city. Le…

Currency Exchange POJ - 1860 题意: 有许多货币兑换点,每个兑换点仅支持两种货币的兑换,兑换有相应的汇率和手续费.你有s这个货币 V 个,问是否能通过合理地兑换货币,使得你手中的货币折合成s后是有增加的. 思路: 这道题在建立每种货币的兑换关系后,找到图中的正环即可,因为你沿着正环跑就可以增加价值.这里可以用类似Bellman_Ford判断负环的方法. #include <algorithm> #include <iterator> #include &…

题目链接:POJ 1860 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points speci…

Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 19881   Accepted: 7114 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and pe…

Currency Exchange Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 60000/30000K (Java/Other) Total Submission(s) : 4   Accepted Submission(s) : 2 Problem Description Several currency exchange points are working in our city. Let us suppose that…

点击打开链接 Currency Exchange Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16635   Accepted: 5821 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies…

Currency Exchange 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/E Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operati…

Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be several points specializing in the…

题目链接: https://cn.vjudge.net/problem/POJ-1860 Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can be s…

题目链接:http://poj.org/problem?id=1860 Description Several currency exchange points are working in our city. Let us suppose that each point specializes in two particular currencies and performs exchange operations only with these currencies. There can b…

DP/单调队列优化 首先不考虑奶牛的喜欢区间,dp方程当然是比较显然的:$ f[i]=min(f[k])+1,i-2*b \leq k \leq i-2*a $  当然这里的$i$和$k$都是偶数啦~这个应该很好理解吧……每次喷灌的都是一个偶数长度的区间嘛…… 那么加上奶牛的喜欢区间的话,只需这样:当$ i>cow[j].x $时,令$ i=cow[j].y , j++$ 也就是说中间的位置全部不考虑放喷灌器. 显然我们对于每个节点的 k 是可以用单调队列维护的!嗯看到这里的同学可以先自己试着去…

可持久化线段树 可持久化线段树是一种神奇的数据结构,它跟我们原来常用的线段树不同,它每次更新是不更改原来数据的,而是新开节点,维护它的历史版本,实现“可持久化”.(当然视情况也会有需要修改的时候) 可持久化线段树的应用有很多,仅以区间第K大这种简单的问题来介绍这种数据结构. 我们原本建立的线段树是表示区间的,或者说,维护的是[位置],存的是每个位置上的各种信息.它的优点是满足区间加法,但不满足区间减法,所以我们这里要换一种建树方式:对于每个区间[1,i]建立一棵权值线段树.这个线段树的作用其实就…

[算法]高斯消元 [题解] 高斯消元经典题型:异或方程组 poj 1222 高斯消元详解 异或相当于相加后mod2 异或方程组就是把加减消元全部改为异或. 异或性质:00 11为假,01 10为真.与1异或取反,与0异或不变. 建图:对于图上每个点x列一条异或方程,未知数为n个灯按不按,系数为灯i按了点x变不变,该行结果n+1为初始状态.(所以a[x][y]其实表示x和y是否存在异或关系) 建图原理见上面链接. 寻找:因题目保证有解,而系数只有0或1,所以不用找最大,找到一个非0系数即可. 消元…

[算法]平衡树(treap) [题解]treap知识见数据结构 在POJ把语言从G++换成C++就过了……??? #include<cstdio> #include<algorithm> #include<ctime> using namespace std; ; ]; int n,m,root,sz,s,tt,d[maxn]; void rr(int &tt)//右旋 { int k=t[tt].l; t[tt].l=t[k].r; t[k].r=tt; tt…

网络流/最大流 愚人节快乐XD 这题是给一个混合图(既有有向边又有无向边),让你判断是否有欧拉回路…… 我们知道如果一个[连通]图中每个节点都满足[入度=出度]那么就一定有欧拉回路…… 那么每条边都可以贡献一个出度出来,对于一条边u->v: 连S->edge cap=1; 如果是有向边,就连 edge->v cap=1; 否则(无向边)连edge->u cap=1, edge->v cap=1; 然后每个点的总度数我们是知道的……那么它最后的[出度]就等于 总度数/2.(这个…

[题目] ExponentiationTime Limit: 500MS Memory Limit: 10000KTotal Submissions: 123707 Accepted: 30202Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the nat…

[算法]矩阵快速幂 [题解] 根据f[n]=f[n-1]+f[n-2],可以构造递推矩阵: $$\begin{vmatrix}1 & 1\\ 1 & 0\end{vmatrix} \times \begin{vmatrix}f_n \\ f_{n-1} \end{vmatrix}=\begin{vmatrix}f_{n+1}\\f_n\end{vmatrix}\\$$ 写成幂形式: $$\begin{vmatrix}1 & 1\\ 1 & 0\end{vmatrix}^n…

[算法]二分+矩阵快速幂 [题意]给定矩阵A和整数k,MOD,求A^0+A^1+A^2+...+A^k. [题解] 定义题目要求的答案为f(n),即: $$f_n=\sum_{i=0}^{n}A^i$$ 当n为偶数时,可以拆成两半,后一半由前一半集体乘A(n/2)得到,即: $$f_n=f_{\frac{n}{2}}(A^{\frac{n}{2}}+1)$$ 当n为奇数时,直接递推: $$f_n=f_{n-1}*A^n$$ 复杂度O(n^3 log k). 快速幂的单位矩阵是主对角线(左上到右下…

[算法]边双连通分量 [题意&题解]http://blog.csdn.net/geniusluzh/article/details/6619575 (注意第一份代码是错误的) 一些细节: 1.判断桥只能在树边判断,不能在反向边判断,体现在程序中注释的wrong位置. 2.标记桥要双向标记. 3.第二次dfs的时候记得不走已染色的点,否则会陷入环中. 核心判断: 1.未访问过:取low 2.未访问过:判桥 3.访问过:取dfn #include<cstdio> #include<a…

[题意]给定a,b,c,在天平左边放置若干重量a的砝码,在天平右边放置若干重量b的砝码,使得天平两端砝码差为c.设放置x个A砝码和y个B砝码,求x+y的最小值. [算法]数论(扩展欧几里德算法) [题解]问题转化为求满足方程ax+by=c,|x|+|y|的最小值. 先用扩展欧几里得算法求得通解. 由原方程得答案分布在y=-a/b*x+c/b(a>0,b>0,c>0),因此是k<0,b>0的直线. 由于斜率一定,min{|x|+|y|}一定出现在x轴两侧或y轴两侧,取这四个点判…