t记录每个格子最早被砸的时间,bfs(x,y,t)表示当前状态为(x,y)格子,时间为t.因为bfs,所以先搜到的t一定小于后搜到的,所以一个格子搜一次就行 #include<iostream> #include<cstdio> #include<queue> using namespace std; const int N=505,inf=1e9,dx[]={-1,1,0,0,0},dy[]={0,0,-1,1,0}; int n,m,t[N][N]; bool v[…
Description CIA headquarter collects data from across the country through its classified network. They have been using optical fibres long before it's been deployed on any civilian projects. However they are still under a lot pressure recently becaus…
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 retur…
BZOJ_3049_[Usaco2013 Jan]Island Travels _状压DP+BFS Description Farmer John has taken the cows to a vacation out on the ocean! The cows are living on N (1 <= N <= 15) islands, which are located on an R x C grid (1 <= R, C <= 50). An island is a…