Caves and Tunnels Time limit: 3.0 secondMemory limit: 64 MB After landing on Mars surface, scientists found a strange system of caves connected by tunnels. So they began to research it using remote controlled robots. It was found out that there exist…

URAL1553 维护一棵树,随时修改某个节点的权值,询问(x,y)路径上权值最大的点. 树是静态的,不过套动态树也能过,时限卡的严就得上树链剖分了. 还是那句话 splay的核心是splay(x) LCT的核心是access(x) 把SPOJ OTOCI的代码改了两行就过了 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespac…

1553. Caves and Tunnels Time limit: 3.0 second Memory limit: 64 MB After landing on Mars surface, scientists found a strange system of caves connected by tunnels. So they began to research it using remote controlled robots. It was found out that ther…

一颗树 每次出发点右键值是0 2操作模式1.第一i右键点值添加x 2.乞讨u至v在这条路上右上方值 树为主的连锁分裂称号 #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 100010; struct edge { int v, next; }e[maxn*2]; int first[maxn], cnt; void AddEdg…

简单题,主要为了练手. #include <cstdio> #include <iostream> #define maxn 100010 using namespace std; namespace L { ], rtg[maxn], val[maxn], mxv[maxn]; inline void update( int nd ) { mxv[nd] = max( val[nd], max(mxv[son[nd][]],mxv[son[nd][]]) ); } void ro…

模板参考:https://blog.csdn.net/saramanda/article/details/55253627 综合各位大大博客后整理的模板: #include<iostream> #include<cstdio> using namespace std; + ; struct LCT { struct node { ]; //父亲(Splay对应的链向上由轻边连着哪个节点).左右儿子 int reverse;//区间反转标记 bool is_root; //是否是所在…

Sorry, but the Android VPN API doesn’t currently allow TAP-based tunnels. Edit .ovpn configfile “dev tun” Tun interface setup failed:tun_prop_error: ifconfig addresses are not in the same /30 subnet (topology net30)…

Tunnels Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 844    Accepted Submission(s): 249 Problem Description Bob is travelling in Xi’an. He finds many secret tunnels beneath the city. In his…

题目链接 The input contains mutiple testcases. Please process till EOF.For each testcase, the first line contains two integers N (1 ≤ N ≤ 15), the side length of the square map and M (1 ≤ M ≤ 15), the number of tunnels.The map of the city is given in the…

Destroy Tunnels 原来早忘记了离散里含有这么一个叫传递闭包的东西 矩阵A的闭包B = A U A^2 U A^3 U ... 所以这里直接如果A[i][j]!= 0,建边i->j跑一遍强连通,看是不是只有一个强连通分量,>=2说明不能所有点都!=0输出exists 否则说明所有i->j(i!=j)都有B[i][j]!= 0输出not exists #include <map> #include <set> #include <stack>…