链接一下题目:luoguP3369[模板]普通平衡树(Treap/SBT) 平衡树解析 #include<iostream> #include<cstdlib> #include<cstdio> #include<cmath> #include<cstring> #include<iomanip> #include<algorithm> #include<ctime> #include<queue>…

题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his barn (), conveniently numbered . Each pipe connects a pair of stalls, and all stalls are connected to each-other via paths of pipes. FJ is pumping milk…

[USACO10MAR]仓配置Barn Allocation 思路: 贪心+线段树维护: 代码: #include <bits/stdc++.h> using namespace std; #define maxn 100005 #define INF 0x3f3f3f3f #define maxtree maxn<<2 struct QueryType { int l,r; bool operator<(const QueryType pos)const { if(pos.…

[ZJOI2015]幻想乡战略游戏 思路: 树剖暴力转移: 代码: #include <bits/stdc++.h> using namespace std; #define maxn 100005 #define ll long long struct TreeNodeType { ll l,r,sum,mid; }; ]; ll n,m,head[maxn],E[maxn<<],V[maxn<<],W[maxn<<],cnt; ll id[maxn],s…

[HNOI2010]BOUNCE 弹飞绵羊 思路: SBlct: 代码: #include <bits/stdc++.h> using namespace std; #define maxn 200005 ],rev[maxn],ki[maxn],sta[maxn],top,lit,size[maxn]; inline void in(int &now) { ; ')Cget=getchar(); ') { now=now*+Cget-'; Cget=getchar(); } } vo…

斐波那契数列 思路: 矩阵快速幂: 来,上代码: #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define mod 1000000007 struct MatrixType { ][]; void mem() { ;i<;i++) { ;j<;j++) ai[i][j]=; } } }; lo…

题目描述 松鼠的新家是一棵树,前几天刚刚装修了新家,新家有n个房间,并且有n-1根树枝连接,每个房间都可以相互到达,且俩个房间之间的路线都是唯一的.天哪,他居然真的住在”树“上. 松鼠想邀请小熊维尼前来参观,并且还指定一份参观指南,他希望维尼能够按照他的指南顺序,先去a1,再去a2,......,最后到an,去参观新家.可是这样会导致维尼重复走很多房间,懒惰的维尼不听地推辞.可是松鼠告诉他,每走到一个房间,他就可以从房间拿一块糖果吃. 维尼是个馋家伙,立马就答应了.现在松鼠希望知道为了保证维尼有…

题目描述 Farmer John always wants his cows to have enough water and thus has made a map of the N (1 <= N <= 700) water pipes on the farm that connect the well to the barn. He was surprised to find a wild mess of different size pipes connected in an appa…

题目描述 Farmer John has N barren pastures (2 <= N <= 100,000) connected by N-1 bidirectional roads, such that there is exactly one path between any two pastures. Bessie, a cow who loves her grazing time, often complains about how there is no grass on t…

题目描述 异或是一种神奇的运算,大部分人把它总结成不进位加法. 在生活中…xor运算也很常见.比如,对于一个问题的回答,是为1,否为0.那么: (A是否是男生 )xor( B是否是男生)＝A和B是否能够成为情侣 好了,现在我们来制造和处理一些复杂的情况.比如我们将给出一颗树,它很高兴自己有N个结点.树的每条边上有一个权值.我们要进行M次询问,对于每次询问,我们想知道某两点之间的路径上所有边权的异或值. 输入输出格式 输入格式: 输入文件第一行包含一个整数N,表示这颗开心的树拥有的结点数,以下有N…