Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37050   Accepted: 16122 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the…

由于解集只为{0, 1, 2}故消元后需dfs枚举求解 #include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath…

题意:给出a和b的gcd和lcm,让你求a和b.按升序输出a和b.若有多组满足条件的a和b,那么输出a+b最小的.思路:lcm=a*b/gcd   lcm/gcd=a/gcd*b/gcd 可知a/gcd与b/gcd互质,由此我们可以先用Pollard_rho法对lcm/gcd进行整数分解, 然后对其因子进行深搜找出符合条件的两个互质的因数,然后再都乘以gcd即为输出答案. #include <iostream> #include <stdio.h> #include <alg…

题意:每组数据给出两行,第一行给出变量,第二行给出约束关系,每个约束包含两个变量x,y,表示x<y.    要求:当x<y时,x排在y前面.让你输出所有满足该约束的有序集. 思路:用拓扑排序,dfs枚举即可,为简便起见,这里将字符变量转化为整型值存储. #include <iostream> #include <stdio.h> #include <string> #include <cstring> #include <algorithm…

想到枚举m个点,然后求最小生成树,ratio即为最小生成树的边权/总的点权.但是怎么枚举这m个点,实在不会.网上查了一下大牛们的解法,用dfs枚举,没想到dfs还有这么个作用. 参考链接:http://blog.csdn.net/xingyeyongheng/article/details/9373271 #include <stdio.h> #include <string.h> #include <set> #include <vector> #incl…

Description Flip game squares. One side of each piece is white and the other one is black and each piece is lying either it's black or white side up. Each round you flip 3 to 5 pieces, thus changing the color of their upper side from black to white a…

题目:http://poj.org/problem?id=1753 这个题在開始接触的训练计划的时候做过,当时用的是DFS遍历,其机制就是把每一个棋子翻一遍.然后顺利的过了.所以也就没有深究. 省赛前一次做PC2遇到了差点儿一模一样的题,仅仅只是是把棋盘的界限由4X4改为了5X5,然后一直跑不出结果来,可是当时崔老湿那个队过了,在最后总结的时候.崔老湿就说和这个题一样,只是要枚举第一行进行优化. 我以为就是恢复第一行然后第二行以此类推,只是手推一下结果是6不是4,就知道这个有问题. 问了崔老湿,…

Fire Net Problem Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small castle that has four openings t…

链接:poj 2965 题意:给定一个4*4矩阵状态,代表门的16个把手.'+'代表关,'-'代表开.当16个把手都为开(即'-')时.门才干打开,问至少要几步门才干打开 改变状态规则:选定16个把手中的随意一个,能够改变其本身以及同行同列的状态(即若为开,则变为关,若为关,则变为开),这一次操作为一步. 分析:这题与poj 1753思路差点儿相同,每一个把手最多改变一次状态, 全部整个矩阵最多改变16次状态 思路:直接dfs枚举全部状态,直到找到目标状态 可是要打印路径,全部应在dfs时记录路…

Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a refrigerator. There are 16 handles on the refrigerator door. Every handle can be in one of two states: open or closed. The refrigerat…