Free Cat~~~ 跳动的喵

POJ - 1637点我点我:-)Sightseeing tourTime Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionThe city executive board in

【分析】最小生成树的唯一性，思路是先判断每条边是否有重边，有的话eq=1，否则0.然后第一次求出最小生成树，将结果记录下来， 然后依次去掉第一次使用过的且含有重边的边，再求一次最小生成树，若结果与第一次结果一样，则不唯一。#include#include#include#include#include#includeusing namespace std;str

POJ - 2831Can We Build This One?Time Limit: 5000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus Description“Highways are built, then l

POJ - 3255点我点我:-)RoadblocksTime Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionBessie has moved to a small fa

POJ - 3463点我点我:-)SightseeingTime Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionTour operator Your Person

POJ - 2449点我点我:-)Remmarguts' DateTime Limit: 4000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus Description"Good man never make

Telephone WireTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3198 Accepted: 1119DescriptionFarmer John's cows are getting restless about their poor t

欧拉函数是指：对于一个正整数n，小于n且和n互质的正整数（包括1）的个数，记作φ(n) 。通式：φ(x)=x*(1-1/p1)*(1-1/p2)*(1-1/p3)*(1-1/p4)…..(1-1/pn),其中p1, p2……pn为x的所有质因数，x是不为0的整数。φ(1)=1（唯一和1互质的数就是1本身）。对于质数p，φ(p) = p - 1。注意φ(1)=1.欧拉定理：对于互质的正整

题目大意:   给出一张无向连通图，求S到E经过k条边的最短路。POJ - 3613点我点我:-)Cow RelaysTime Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus Descripti

Harmony ForeverTime Limit: 10000MS Memory Limit: 65536KTotal Submissions: 5834 Accepted: 1414DescriptionWe believe that every inhabitant of this universe eventual

（ＩＤＡ＊：　即为A*算法加上迭代加深搜索，本题为IDA*）DescriptionThe Eastowner city is perpetually haunted with water supply shortages, so in order to remedy this problem a new water-pipe has been built. Builders

POJ - 1155点我点我:-)TELETime Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionA TV-network plans to broadcast an impor

POJ - 2914点我点我Minimum CutTime Limit: 10000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionGiven an undirected graph, in whi

题目大意：给定一个有向图，根节点已知，求该有向图的最小树形图。最小树形图即有向图的最小生成树，定义为：选择一些边，使得根节点能够到达图中所有的节点，并使得选出的边的边权和最小。题目算法：朱-刘算法（即由中国人朱永津和刘振宏共同发明的算法）。算法步骤如下：（本文不再证明，参考下面给出的我自己画的一个图即可理解）1.判断图的连通性，若不连通直接无解，否则一定有解。2.为

POJ - 2989点我点我All FriendsTime Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionSociologists are interested in the p

POJ - 3621点我点我:-)Sightseeing CowsTime Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionFarmer John has decided

POJ - 2584点我点我:-)T-Shirt GumboTime Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionBoudreaux and Thibodeaux are stud

POJ - 1019点我点我:-)大致题意：有一串数字串，其规律为1 12 123 1234 12345 123456 1234567 12345678 123456789 12345678910 1234567891011 123456789101112······k输入位置n，计算这一串数字第n位是什么数字，注意是数字，不是数！例如12345678910的第10位是

poj 1306  Combinations       下面给出两种方法^-^(都AC!!!)#include#include#include#include#include#includeusing namespace std;#define For(i, a, b) for(int i = a; i <= b; i++)#define ULL

POJ - 1131Octal FractionsTime Limit: 1000MS Memory Limit: 10000KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionFractions in octal (base 8) notation

POJ - 2135点我点我:-)Farm TourTime Limit: 1000MS Memory Limit: 65536KB 64bit IO Format: %lld & %lluSubmitStatus DescriptionWhen FJ's friends visit him on the

Solitaire is a game played on a chessboard 8x8. The rows and columns of the chessboard are numbered from 1 to 8, from the top to the bottom and from left to right respectively.There are four ide