POJ 1470 Closest Common Ancestors
2024-09-29 22:34:59
Closest Common Ancestors
Description Write a program that takes as input a rooted tree and a list of pairs of vertices. For each pair (u,v) the program determines the closest common ancestor of u and v in the tree. The closest common ancestor of two nodes u and v is the node w that is an ancestor of both u and v and has the greatest depth in the tree. A node can be its own ancestor (for example in Figure 1 the ancestors of node 2 are 2 and 5)
Input The data set, which is read from a the std input, starts with the tree description, in the form:
nr_of_vertices The input file contents several data sets (at least one). Output For each common ancestor the program prints the ancestor and the number of pair for which it is an ancestor. The results are printed on the standard output on separate lines, in to the ascending order of the vertices, in the format: ancestor:times
For example, for the following tree: Sample Input 5 Sample Output 2:1 Hint Huge input, scanf is recommended.
Source |
-----------------------------------------------------------------------
LCA
采用 Tarjan 离线 LCA 算法比较方便
注意读入细节
-------------------------------------------------------------------------
#include <cstdio>
#include <vector>
#include <cstring>
#define pb push_back using namespace std;
const int N();
vector<int> q[N], g[N];
int par[N], ans[N], col[N];
int find(int u){return par[u]==u?u:find(par[u]);}
void dfs(int u, int f){
col[u]=-;
for(int i=; i<q[u].size(); i++){
int &v=q[u][i];
if(col[v]==-) ans[v]++;
else if(col[v]==) ans[find(v)]++;
else q[v].pb(u);
}
for(int i=; i<g[u].size(); i++){
int &v=g[u][i];
dfs(v, u);
}
col[u]=;
par[u]=f;
}
int main(){
//freopen("in", "r", stdin);
int n, m, u, v;
for(;~scanf("%d", &n);){
for(int i=; i<=n; i++) g[i].clear(), q[i].clear();
memset(par, , sizeof(par));
for(int i=; i<n; i++){
scanf("%d:(%d)", &u, &m);
while(m--){
scanf("%d", &v);
par[v]=u;
g[u].pb(v);
}
}
scanf("%d", &m);
while(m--){
scanf(" (%d%d)", &u, &v);
q[u].pb(v);
}
int rt;
for(rt=; par[rt]; rt=par[rt]);
for(int i=; i<=n; i++) par[i]=i;
memset(ans, , sizeof(ans));
memset(col, , sizeof(col));
dfs(rt, rt);
for(int i=; i<=n; i++) if(ans[i]) printf("%d:%d\n", i, ans[i]);
}
}
最新文章
- Android项目实战(二十五):Android studio 混淆+打包+验证是否成功
- git之四
- iOS-微信支付(订单号重复的问题)
- 搭建HTTP Live Streaming直播系统
- EasyUI之DataGrid使用
- excel 组及分级显示制作教程
- ZOJ 3941 Kpop Music Party 贪心
- 06.Hibernate实体类生命周期
- VirtualBox扩展磁盘空间
- oracle琐碎笔记
- Android:Asmack能登录但是获取不到联系人的问题
- Scheme实现二叉查找树及基本操作(添加、删除、并、交)
- web端常见安全漏洞测试结果分析-- appscan
- Angualr学习笔记
- org.hibernate.ObjectNotFoundException: No row with the given identifier exists解决办法
- Python代码编写规范
- POJ1569 Myacm Triangles
- jdk1.7安装,cmd下 java -version出现错误:“could not open `D:\Java\jre7\lib\amd64\jvm.cfg”
- 使用poi导出Excel,并设定单元格内容类型,抛出异常
- 神经网络损失函数中的正则化项L1和L2