题目: 链接:点击打开链接题意:思路: 一个并查集,题目就是要让你判断是否是一个连通的无环图。1>判断成环的时候,只要判断输入边的两个点。有一个共同的
https://www.u72.net/daima/zms5.html - 2024-07-05 10:55:20 - 代码库并查集的应用。实质上是判断这是否是一棵树。需要注意的是0 0 也是一棵树。#include<cstdio>#include<cstring>#include<algorithm>using namespac
https://www.u72.net/daima/cr2b.html - 2024-07-10 23:32:57 - 代码库题目链接:http://acm.nyist.net/JudgeOnline/problem.php?pid=82 做这道题的时候迷迷糊糊的,,果然比较难。。最后也是没有做出来。。请教了一下学长,学长
https://www.u72.net/daima/xf9m.html - 2024-07-17 00:21:19 - 代码库本题可以使用BFS和DFS解题,也可以构建图,然后利用Dijsktra解题。不过因为数据很少,就没必要使用Dijsktra了。BFS和DFS效率都是一样的,因为都需要搜索所有可
https://www.u72.net/daima/xdr8.html - 2024-07-16 22:00:46 - 代码库#include <iostream> #include <iomanip> #include <cstdlib> using namespace std; #define MaxSize 100 int maze[10][10] = //定
https://www.u72.net/daima/x7s9.html - 2024-08-27 22:28:10 - 代码库#include<stdio.h>#include<string.h>#include<algorithm>#include<stack>using namespace std;int a[5][5],b[5][5];int di[4][2]={0,1,0,-1,1
https://www.u72.net/daima/62xx.html - 2024-09-08 18:19:23 - 代码库http://cpp.zjut.edu.cn/ShowProblem.aspx?ShowID=1423设dp[i]表示在i点时到达终点要走的期望步数,那么dp[i] = ∑1/m*dp[j] &#43; 1,j是与i相连的点,m是
https://www.u72.net/daima/5b1d.html - 2024-07-23 02:37:39 - 代码库There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up (u), down (d), left (l) or right
https://www.u72.net/daima/mzez.html - 2024-09-16 09:44:11 - 代码库记录路径问题和层数利用结构体变量可以很容易的实现 这里要求路劲字典序列最小 改变一下优先的方向就可以了(注意实际方向和二维数组的对应关系,坑死我了
https://www.u72.net/daima/82k6.html - 2024-09-12 03:38:44 - 代码库http://yuncode.net/code/c_5093587f5dbaf9该文章使用递归可以寻找出路径,但并未寻找出所有的路径。#include <stdio.h>#include <stdlib.h>void vi
https://www.u72.net/daima/e4wk.html - 2024-07-28 19:38:33 - 代码库There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won
https://www.u72.net/daima/e3ks.html - 2024-09-15 15:48:28 - 代码库1 .Preface/*** There have been many data to introduce the algorithm. So I will try to simply explain it and explain the program in detail.
https://www.u72.net/daima/b44n.html - 2024-07-09 07:37:05 - 代码库Tarjan算法解读:https://www.byvoid.com/zht/blog/scc-tarjan 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<cstrin
https://www.u72.net/daima/nzbsk.html - 2024-09-21 20:24:44 - 代码库http://acm.hdu.edu.cn/showproblem.php?pid=1269题意:确定是否是一个强连通图。思路:裸的tarjan算法。 1 #include <cstdio> 2 #include <algorithm> 3
https://www.u72.net/daima/d18v.html - 2024-08-15 08:01:34 - 代码库#include <cstdio>#include <cmath>#include <iostream>#include <algorithm>#include <cstdlib>#include <cstring>#include <map>#include <v
https://www.u72.net/daima/kc7f.html - 2024-07-06 21:19:29 - 代码库Description定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0,};它表示一
https://www.u72.net/daima/w5ur.html - 2024-07-16 11:00:51 - 代码库题目传送门 基础的并查集,稍微改下模板就可以。 要注意两个条件: ①判断合并的两个点根节点是否相同 ②每个点能否全部连通
https://www.u72.net/daima/21uk.html - 2024-09-01 15:39:07 - 代码库宽度优先搜索按照距离开始状态由近及远的顺序进行搜索,可以很容易用来求解最短路径或者最少操作等问题。将已经访问过的状态用标记管理起来,便可以很好
https://www.u72.net/daima/8szd.html - 2024-09-11 16:52:06 - 代码库import random#warning: x and y confusingsx = 10sy = 10dfs = [[0 for col in range(sx)] for row in range(sy)]maze = [[&#39; &#39; for c
https://www.u72.net/daima/nvkd0.html - 2024-10-28 14:33:40 - 代码库Description 定义一个二维数组: int maze[5][5] = { 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0,};它表示
https://www.u72.net/daima/nc9dz.html - 2024-08-08 18:24:45 - 代码库