zhx and contest (枚举  + dfs)】的更多相关文章

zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 324    Accepted Submission(s): 118 Problem Description As one of the most powerful brushes in the world, zhx usually takes part i…
题目:http://poj.org/problem?id=2965 来源:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26732#problem/B 题意: 就是把题目中给出的状态图,全部翻转成 ---- ---- ---- ----状态 翻转: 每次翻转一个,那么它所在的行和列都要翻转 问最小翻转次数,同时输出翻转路径. 算法: 暴力 + 枚举 + dfs 思路: 可以证明每个把手要么翻转,要么不翻转,那么从左到右,从上到下依次枚…
传送门 zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 145    Accepted Submission(s): 49 Problem Description As one of the most powerful brushes in the world, zhx usually takes part…
zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 851    Accepted Submission(s): 282 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n p…
zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 455    Accepted Submission(s): 158 Problem Description As one of the most powerful brushes in the world, zhx usually takes part i…
传送门 zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 575    Accepted Submission(s): 181 Problem Description As one of the most powerful brushes, zhx is required to give his juniors…
zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3835    Accepted Submission(s): 1255 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n…
又一发吐血ac,,,再次明白了用函数(代码重用)和思路清晰的重要性. 11779687 2014-10-02 20:57:53 Accepted 4770 0MS 496K 2976 B G++ czy Lights Against Dudely Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1360    Accepted Subm…
Legendary Teams Contest Time limit: 1.0 secondMemory limit: 64 MB Nothing makes as old as years. A lot of cool contests are gone, a lot of programmers are not students anymore and are not allowed to take part at the contests. Though their spirit is f…
题目链接:http://poj.org/problem?id=3740 题意: 是否从0,1矩阵中选出若干行,使得新的矩阵每一列有且仅有一个1? 原矩阵N*M $ 1<= N <= 16 $ , $ 1 <= M <= 300$ 解法1:由于行数不多,二进制枚举选出的行数,时间复杂度为O((1<<16)*K), 其中K即为判断选出的行数是否存在相同的列中有重复的1: 优化:将1状压即可,这样300的列值,压缩在int的32位中,使得列数“好像”小于10了:这样每次只需要…
Problem UVA818-Cutting Chains Accept:393  Submit:2087 Time Limit: 3000 mSec  Problem Description What a find! Anna Locke has just bought several links of chain some of which may be connected. They are made from zorkium, a material that was frequently…
Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 40632   Accepted: 17647 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…
思路: 枚举+搜一下+判个重 ==AC //By SiriusRen #include <set> #include <cstdio> using namespace std; int a[8][8],xx[]={1,-1,0,0},yy[]={0,0,1,-1}; set<int>s; bool check(int x,int y){ return x>0&&x<6&&y>0&&y<6; } vo…
Problem Description As one of the most powerful brushes in the world, zhx usually takes part in all kinds of contests. One day, zhx takes part in an contest. He found the contest very easy for him. There are n problems in the contest. He knows that h…
Problem Description Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of whi…
题意: 给你一个由n个点,n-1条有向边构成的一颗树,1为根节点 下面会输入n-1个数,第i个数表示第i+1点的父节点.你可以去添加一条边(你添加的边也是有向边),然后找出来(x,y)这样的成对节点.问你最多能找出来多少对 其中x和y可以相等,且x点要可以到达y点 题解: 根据样例找一下就可以看出来让根节点1和深度最深那个点相连之后能找出来的(x,y)最多 但是又出现一个问题,如果那个最大深度的点不止一个,那么我们要选择那个.如下样例 6 1 1 2 2 3 化成图就是 4.5.6号点都是最深深…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4770 思路:由于最多只有15个".",可以直接枚举放置的位置,然后判断是否能够全部点亮即可.需要注意的是,有一个特殊的light,也需要枚举它的位置以及放置的方向. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using names…
The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22057   Accepted: 8521   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…
题目描述 FJ的N(1 <= N <= 100)头奶牛们最近参加了场程序设计竞赛:).在赛场上,奶牛们按1..N依次编号.每头奶牛的编程能力不尽相同,并且没有哪两头奶牛的水平不相上下,也就是说,奶牛们的编程能力有明确的排名. 整个比赛被分成了若干轮,每一轮是两头指定编号的奶牛的对决.如果编号为A的奶牛的编程能力强于编号为B的奶牛(1 <= A <= N; 1 <= B <= N; A != B) ,那么她们的对决中,编号为A的奶牛总是能胜出. FJ想知道奶牛们编程能力的…
Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11129   Accepted: 6183 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than other…
[问题描述] 作为史上最强的刷子之一,zhx的老师让他给学弟(mei)们出n道题.zhx认为第i道题的难度就是i.他想要让这些题目排列起来很漂亮. zhx认为一个漂亮的序列{ai}下列两个条件均需满足. 1:a1..ai是单调递减或者单调递增的. 2:ai..an是单调递减或者单调递增的. 他想你告诉他有多少种排列是漂亮的.因为答案很大,所以只需要输出答案模p之后的值. [题解] 对于整段序列单调增或减,就2种情况. 对于先增后减或先减后增:确定一个序列中的最小值(最大值),其他的要么在左边,要…
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2388    Accepted Submission(s): 755 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n problems.zhx t…
题目链接:pid=4462">传送门 题意:一个n*n的区域,有m个位置是能够放稻草人的.其余都是玉米.对于每一个位置(x,y)所放稻草人都有个作用范围ri, 即abs(x-i)+abs(y-j)<=r,(i,j)为作用范围内.问至少要在几个位置上放稻草人,才干覆盖全部的玉米,若不可能则输出-1. 有一个trick,就是放稻草人的位置不用被覆盖 eg: input: 2 4 1 1 1 2 2 1 2 2 0 0 0 0 output: 0 0 代码例如以下: #include &l…
作为史上最强的刷子之一,zhx的老师让他给学弟(mei)们出n道题.zhx认为第i道题的难度就是i.他想要让这些题目排列起来很漂亮. zhx认为一个漂亮的序列{ai}下列两个条件均需满足. 1:a1..ai是单调递减或者单调递增的. 2:ai..an是单调递减或者单调递增的. 他想你告诉他有多少种排列是漂亮的.因为答案很大,所以只需要输出答案模p之后的值. Input Multiply test cases(less than 10001000). Seek EOF as the end of…
题目分析如果n=1,答案是1,否则答案是2n−2. 证明:ai肯定是最小的或者最大的.考虑另外的数,如果它们的位置定了的话,那么整个序列是唯一的. 那么ai是最小或者最大分别有2n−1种情况,而整个序列单调增或者单调减的情况被算了2次,所以要减2. 要注意的一点是因为p>231,所以要用快速乘法.用法与快速幂相同.如果直接乘会超过long long范围,从而wa掉. 快速幂+快速乘法 #include<cstdio>#include<cstring>#include<c…
题目链接: hdu: http://acm.hdu.edu.cn/showproblem.php?pid=5187 bc(中文): http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=571&pid=1002 题解: 求(2^n-2)%p,题目看错,一天都没什么思路,冷静一下.. 代码: #include<iostream> #include<cstring> #include<cst…
求确定身份的人的个数. 只能确定狼的身份,因为只能找到谁说了谎.但一个人是否是民,无法确定. 将人视作点,指认关系视作边,有狼边和民边两种边. 确定狼的方法只有两种: 1. 在一个仅由一条狼边组成的环中,狼边指向的那个点必定是狼. 2. 环外指认铁狼为民的也必定是狼. 所以用原图找环求情况1中的铁狼,反向建图找情况2中的狼. #include <bits/stdc++.h> using namespace std; typedef long long LL; ; const int INF =…
zhx's contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 448    Accepted Submission(s): 147 Problem Description As one of the most powerful brushes, zhx is required to give his juniors n p…
  N皇后问题 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 12279    Accepted Submission(s): 5535 Problem Description 在N*N的方格棋盘放置了N个皇后,使得它们不相互攻击(即任意2个皇后不允许处在同一排,同一列,也不允许处在与棋盘边框成45角的斜线上.你的任务是,对于给定的N,…
http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=32 组合数 时间限制:3000 ms  |  内存限制:65535 KB 难度:3   描述 找出从自然数1.2.... .n(0<n<10)中任取r(0<r<=n)个数的所有组合.   输入 输入n.r. 输出 按特定顺序输出所有组合.特定顺序:每一个组合中的值从大到小排列,组合之间按逆字典序排列. 样例输入 5 3 样例输出 543 542 541 532 531 521…