命运在指间

题目描述给定一个无向连通图，顶点编号从0到n-1，用广度优先搜索(BFS)遍历，输出从某个顶点出发的遍历序列。(同一个结点的同层邻接点，节点编号小的优先遍历）输入输入第一行为整数n（0对于每组数据，第一行是三个整数k，m，t（0＜k＜100，0＜m＜(k-1)*k/2，0＜ t＜k），表示有m条边，k个顶点，t为遍历的起始顶点。下面的m行，每行是空格隔开的两个整数u，v，表示一

Time Limit: 2000ms   Memory limit: 65536K  有疑问？点这里^_^题目描述Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,0

Nemo is a naughty boy. One day he went into the deep sea all by himself. Unfortunately, he became lost and couldn't find his way home. Therefore, he sent a signal to his father, Marlin, to ask for hel

题意：从S到E，一个三维的BFS可以解决，多一维，可以这样去理解，二维相当于一个平面，而三维的相当于一栋楼房，只是在z上多了几个平面而已另外，第一次用队列写BFS好开心，以前都是模拟队列，也是醉了16ms#include#include#include#include#include#includeusing namespace std;int dx[]= {

题意：给你一个数n，让你找到一个由0,1构成的n的倍数思路：姿势对了，就过了1.0#include#include#include#include#include#includeusing namespace std;long long bfs(int n){ queuep; while(!p.empty()) { p.pop

题意： 给你两个素数，n,m； 问你能够最少几步把n变成m； 变的规则： 每次能够变一位数字，要求变完，仍然是素数； n,m都是4位数思路： BFS，先用素数筛把10000都筛出来，然后用BFS，对个十百千位的数字去枚举（千位不能是0）16ms#include#include#include#include#include#includeusing