简介:Find The Multiple(dfs和bfs都可) Given a positive integer n, write a program to find out a nonzero multiple m of n whose decimal representation contains only the digits 0 and 1. You may assume that n is not greater than 200 and there is a corresponding m containing no more than ...
解法一,DFS 深度优先搜索 解法二,BFS 广度搜索优先(DFS+临时队列) BFS 的 解法三,UnionFind 并查集(Disjoin Set) Find 函数的解释 Leetcode 新手快速上手100题代码整理:王几行xing:LeetCode 力扣入门100题 (全网新手最友好!) 本体涉及的数据结构:图,或者简单而言,叫二维数组 读题 关键:只考虑上下左右的方向,...
两种解法:DFS和BFS都行。 DFS: #include<iostream>usingnamespacestd;intn;boolflag;voiddfs(longlongx,intk) {if(flag)return;if(x%n==0) { flag=true; cout<<x<<endl;return; }if(k==18)//搜索到18位的时候还没找到就返回return; dfs(x*10,k+1); dfs(x*10+1,k+1); }intmain() {whil...
1/*2题意:找出一个0和1组成的数字能整除n3DFS:200的范围内不会爆long long,DFS水过~4*/5/***6Author :Running_Time7Created Time :2015-8-2 14:21:518File Name :POJ_1426.cpp9***/1011#include <cstdio>12#include <algorithm>13#include <iostream>14#include <sstream>15#include <cstring>16#...
Java uses JGraphT's DFS traversal, Disjoint-set data structure and other data structures to find the connectivity of the graph
The graph is acyclic. It's not a distributed system so you have full control (obviously ...). Because you don't care about path and because the graph is acyclic, the table on each node can be a mapmarked_node_id -> countwhere count is the number of paths from the given node to...
DFS/BFS(同余模) POJ 1426 Find The Multiple 题目传送门 1 /* 2 题意:找出一个0和1组成的数字能整除n 3 DFS:200的范围内不会爆long long,DFS水过~ 4 */ 5 /*** 6 Author :Running_Time 7 Created Time :2015-8-2 14:21:51 8 File Name :POJ_1426.cpp...
The solution is to install: A E B Y. Here is an image to describe the example: Is there an algorithm to solve the problem without using a brute-force approach? I've already read a lot about algorithms such as DFS, BFS, Dijkstra, etc... The problem is that these algorithms ...
今天学习了神奇的并查集,UnionFind,试着做了几道之前用BFS/DFS的算法题,感觉超好用! UnionFind的基础运用: 创建一个UnionFind 查找某元素所在group的大哥 连接两个元素 判断两个元素是否在同一个group中 那我们来看看题目吧! Connecting Graph Given n nodes in a graph labeled from 1 to n. There is no ...
Find The Multiple POJ - 1426 原题链接 考察:dfs或者bfs 暴力枚举即可,没有什么特别的技巧 dfs:找到了就标记一下,避免继续搜索(因为答案一定在long long里,所以超过18位就可以不用算了) bfs:用G++编译,找到当即退出.注意一定要让所有路径都有返回值否则报错...