题目:Nearest Common Ancestors 根据输入建立树,然后求2个结点的最近共同祖先. 注意几点: (1)记录每个结点的父亲,比较层级时要用: (2)记录层级: (3)记录每个结点的孩子,vector<int> v[M]写在主函数里面,放在外面超时. 代码: #include<iostream> #include<vector> #include<cstdio> #include<cstdlib> #include<cstr…

POJ 1330 Nearest Common Ancestors A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, each node is labeled with an integer from {1, 2,...,16}. Node 8 is the root of the tree. No…

POJ - 1330 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000KB   64bit IO Format: %lld & %llu Submit Status Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In t…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30147   Accepted: 15413 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, eac…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14902   Accepted: 7963 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:  In the figure, each…

传送门 Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26612   Accepted: 13734 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below:    In the figur…

Nearest Common Ancestors Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 39596   Accepted: 19628 Description A rooted tree is a well-known data structure in computer science and engineering. An example is shown below: In the figure, each…

多校第七场考了一道lca,那么就挑一道水题学习一下吧= = 最简单暴力的方法:建好树后,输入询问的点u,v,先把u全部的祖先标记掉,然后沿着v->rt(根)的顺序检查,第一个被u标记的点即为u,v的公共祖先. 标记的时候又犯老毛病了:while,do while都不对,最后还是while(1)了T^T #include<cstdio> #include<cstring> ; int p[MAXN],vis[MAXN]; int main() { int T,n,u,v; sc…

1.输入树中的节点数N,输入树中的N-1条边.最后输入2个点,输出它们的最近公共祖先. 2.裸的最近公共祖先. 3. dfs+ST在线算法: /* LCA(POJ 1330) 在线算法 DFS+ST */ #include<iostream> #include<stdio.h> #include<string.h> using namespace std; ; *MAXN];//rmq数组,就是欧拉序列对应的深度序列 struct ST{ *MAXN]; *MAXN][…

详细讲解见:https://blog.csdn.net/liangzhaoyang1/article/details/52549822 zz:https://www.cnblogs.com/kuangbin/p/3302493.html /* *********************************************** Author :kuangbin Created Time :2013-9-5 0:09:55 File Name :F:\2013ACM练习\专题学习\LCA…