Hadis_yuki

Pushing BoxesTime Limit: 2000MS Memory Limit: 131072KTotal Submissions: 3975 Accepted: 1403 Special JudgeDescriptionImagine you are standing inside a two-dime

NightmareTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5475    Accepted Submission(s): 2725Problem DescriptionIgnatius had a night

Red and BlackTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6453    Accepted Submission(s): 4081Problem DescriptionThere is a recta

Flip GameTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 25323 Accepted: 10927DescriptionFlip game is played on a rectangular 4x4 field with two-sided pie

The Pilots Brothers' refrigeratorTime Limit: 1000MS Memory Limit: 65536KTotal Submissions: 15572 Accepted: 5856 Special JudgeDescriptionThe game “The Pilots B

King Arthur's KnightsTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1518 Accepted Submission(s): 645Special JudgeProblem DescriptionI

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4490Mad VeterinarianTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 279    Accepted Submiss

The RobberyTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 1543 Accepted: 151DescriptionIn the downtown of Bucharest there is a very big bank with a very

思路：把出度大于1的点去掉，以每一个hotel为起点搜索找最长路径。链接：点击打开链接#include#include#includeusing namespace std;int len,fa[100010],sum[100010],maxpath,hotel[100010];int dfs(int x){ if(fa[x]==0) r

题意：求给定数n的最小的倍数，使其中不包含给出的m个十进制数。状态太多了。。。参看别人的思路，就是在每一位上对可以出现的数进行枚举，然后bfs。。。直到出现第一个节点中m==0，就输出。每个节点记录三个值c：末尾的那个数字m：这个节点所代表的数字对n取模的值f：父节点对于一个节点，如果在其后面加一个数字i，那么新的m值=(m*10 + i

分析：1、每格棋子最多只可以翻转一次（实际是奇数次，但这没意义），只要其中一格重复翻了2次(不论是连续翻动还是不连翻动)，那么它以及周边的棋子和没翻动时的状态是   一致的，由此就可以确定这个棋盘最多只能走16步，最多只能有翻出2^16种状态           2、翻与不翻，分别向下搜索。就像那个在一列数中找和为m的排列，每次数都有取与不取两种可能，取并向下搜索，不取向下搜索，如果找到就返回

Knight MovesTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4983    Accepted Submission(s): 3046Problem DescriptionA friend of you i

Asteroids!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2599    Accepted Submission(s): 1745Problem DescriptionYou're in space.Y

Finding NemoTime Limit: 2000MS Memory Limit: 30000KTotal Submissions: 6869 Accepted: 1571DescriptionNemo is a naughty boy. One day he went into the deep sea all by

DescriptionLittle Tom loves playing games. One day he downloads a little computer game called 'Bloxorz' which makes him excited. It's a game about rolling a box to a specific position on a special

RescueTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11533    Accepted Submission(s): 4197Problem DescriptionAngel was caught by th

Holedox MovingTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 12103 Accepted: 2899DescriptionDuring winter, the most hungry and severe time, Holedox sleep

胜利大逃亡Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21449    Accepted Submission(s): 8442Problem DescriptionIgnatius被魔王抓走了,有一天魔王出差去

Prime Ring ProblemTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20105    Accepted Submission(s): 9001Problem DescriptionA ring is

A strange liftTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8550    Accepted Submission(s): 3241Problem DescriptionThere is a stra

Oil DepositsTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7813    Accepted Submission(s): 4583Problem DescriptionThe GeoSurvComp g

Tempter of the BoneTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 52416    Accepted Submission(s): 14094Problem DescriptionThe dogg

题意：在迷宫中找存在的外星人A撒。。开始是一个组织找，后来这个组织可以分裂成            很多组。求找到所有外星人的最小步数。（分裂后，步数计算为所有组织的步数和）思路：因为要保证每个外星人A都找到，加上组织所在起始点S，求最小步数。即求图中           n个点的最小生成树（n=外星人的个数+1）。            bfs找每两点间的距离。