Description Input is the matrix A of N by N non-negative integers. A distance between two elements Aij and Apq is defined as |i − p| + |j − q|. Your program must replace each zero element in the matrix with the nearest non-zero one. If there are two…

Nearest number - 2 Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 4100 Accepted: 1275 Description Input is the matrix A of N by N non-negative integers. A distance between two elements Aij and Apq is defined as |i − p| + |j − q|. Your pro…

Open the Lock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3847    Accepted Submission(s): 1661 Problem Description Now an emergent task for you is to open a password lock. The password is co…

Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Ve…

题目链接: http://www.lydsy.com/JudgeOnline/problem.php?id=1295 题解: 对每个点暴力跑一遍bfs,看能够到达的最远位置,这里如果有障碍物则距离为1,如果没有障碍物,则距离为0,用bfs跑距离<=t的所有点并更新答案. 代码: #include<iostream> #include<cstring> #include<cstdio> #include<utility> #include<queu…

Easy Number Challenge Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 236B Appoint description:  System Crawler  (2016-04-26) Description Let's denote d(n) as the number of divisors of a…

题意:给定4个数,a,b,c,d,分别代表空杯子容积为a,b,一个盛满水的杯子容积为c,让你不断倒水,找一个dd,是不是存在某个时刻, 某个杯子里的水dd,和d相同,或者无限接近.让求最少的倒水量和dd(可能和d相同). 析:首先由于没有刻度,如果用数学方法计算,不好算,样例还好算一点,我们观察那个a,b,c都不大于200,挺小的,适合暴力求解. 就是把所有情况都倒一次,倒水就两种倒法,要么把一个杯子倒满,要么就是这个杯子空了,由于水量是固定的,那么确定两个杯子的水量, 那么第三个也就确定了,所…

Description Harry: "But Hagrid. How am I going to pay for all of this? I haven't any money." Hagrid: "Well there's your money, Harry! Gringotts, the wizard bank! Ain't no safer place. Not one. Except perhaps Hogwarts." ― Rubeus Hagrid…

八数码问题是一个经典的人工智能问题.具体问题不累述了. 思路:由于存在多组测试数据,可以考虑“打表法“.所谓打表法,即枚举所有的初始情况,记录其到达终点的路径.而在这个题目中,顺序打表会调用很多次BFS,所以我们采用逆序打表,只要调用一次BFS. 代码如下: /*************map存路径,set判重,string存状态*****************/ /*********************暴力广搜 + STL **************************/ #inc…

题目链接(洛谷) 题目大意 给定两个数 \(u\) , \(v\) .有三种操作: \(u=u+1(mod\) \(p)\) . \(u=u+p−1(mod\) \(p)\) . \(u=u^{p−2}(mod\) \(p)\) . 思路 BFS 状态太多导致队列装不下. 迭代加深 \(TLE\) ,浪费了太多时间在每一层上,没有记忆化且状态很多. IDA* 不行,无法得出乐观股价函数. 双向BFS 这样会将步数很为两半,状态相较于普通的 \(BFS\) 会少很多. 先来看操作一和操作二,他们的…