1450:[例 3]Knight Moves 题解 这道题可以用双向宽度搜索优化(总介绍在 BFS ) 给定了起始状态和结束状态,求最少步数,显然是用BFS,为了节省时间,选择双向BFS. 双向BFS,即从起点向终点搜,从终点向起点搜,扩展各自的状态,直到出现两者扩展的状态重合 优化:每次选择结点少的扩展 看一下骑士可以到达那些点呢?? 所以当然要开两个队列啦 设定: 1. dis[ i ][ a ][ b ]:队列 i ,从起点(a,b)至少多少步 2. v[ i ][ a ][ b…
time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output Kleofáš is participating in an n-thlon - a tournament consisting of n different competitions in n different disciplines (numbered 1 through n). T…
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output The country Treeland consists of n cities connected with n - 1 bidirectional roads in such a way that it's possible to reach every city starting…