题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3085 Nightmare Ⅱ Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2790 Accepted Submission(s): 781 Problem Description Last night, little erriyu
题目传送门 题意:一个人去救女朋友,两个人都在运动,还有鬼在"扩散",问最少几秒救到女朋友 分析:开两个队列来表示两个人走过的路,一个人走到的地方另一个人已经vis了,那么就是相遇了,鬼就用曼哈顿距离判断. #include <bits/stdc++.h> using namespace std; const int N = 8e2 + 5; char maze[N][N]; int n, m; bool vis[N][N][2]; int dx[4] = {-1, 1,
Description In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don't overlap, or at least that they don't conflict. One way of accomplishing this is to restrict a transmitter's cov