Closest Common Ancestors

Time Limit: 2000ms
Memory Limit: 10000KB

This problem will be judged on PKU. Original ID: 1470
64-bit integer IO format: %lld      Java class name: Main

 
 
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 
vertex:(nr_of_successors) successor1 successor2 ... successorn 
...
where vertices are represented as integers from 1 to n ( n <= 900 ). The tree description is followed by a list of pairs of vertices, in the form: 
nr_of_pairs 
(u v) (x y) ...

The input file contents several data sets (at least one). 
Note that white-spaces (tabs, spaces and line breaks) can be used freely in the input.

 

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
5:(3) 1 4 2
1:(0)
4:(0)
2:(1) 3
3:(0)
6
(1 5) (1 4) (4 2)
(2 3)
(1 3) (4 3)

Sample Output

2:1
5:5

Hint

Huge input, scanf is recommended.

 

Source

 
解题:LCA。。。。
 
 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <climits>
#include <algorithm>
#include <cmath>
#define LL long long
#define INF 0x3f3f3f
using namespace std;
const int maxn = ;
vector<int>g[maxn];
vector<int>q[maxn];
int n,m,cnt[maxn],uf[maxn];
bool vis[maxn],indeg[maxn];
int Find(int x) {
if(x != uf[x])
uf[x] = Find(uf[x]);
return uf[x];
}
void tarjan(int u) {
int i;
uf[u] = u;
for(i = ; i < g[u].size(); i++) {
if(!vis[g[u][i]] && g[u][i] != u) {
tarjan(g[u][i]);
uf[g[u][i]] = u;
}
}
vis[u] = true;
for(i = ; i < q[u].size(); i++) {
if(vis[q[u][i]]) cnt[Find(q[u][i])]++;
}
}
int main() {
int i,j,u,v,k;
while(~scanf("%d",&n)) {
for(i = ; i <= n; i++) {
g[i].clear();
q[i].clear();
cnt[i] = ;
indeg[i] = false;
}
for(i = ; i < n; i++) {
scanf("%d:(%d)",&u,&k);
for(j = ; j < k; j++) {
scanf("%d",&v);
g[u].push_back(v);
indeg[v] = true;
}
}
scanf("%d",&m);
while(m--) {
scanf(" (%d %d)",&u,&v);
q[u].push_back(v);
q[v].push_back(u);
}
memset(vis,false,sizeof(vis));
memset(cnt,,sizeof(cnt));
for(i = ; i <= n; i++)
if(!indeg[i]) {
tarjan(i);
break;
}
for(i = ; i <= n; i++)
if(cnt[i]) printf("%d:%d\n",i,cnt[i]);
}
return ;
}

BNUOJ 1589 Closest Common Ancestors的更多相关文章

  1. POJ 1470 Closest Common Ancestors

    传送门 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 17306   Ac ...

  2. poj----(1470)Closest Common Ancestors(LCA)

    Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 15446   Accept ...

  3. POJ 1470 Closest Common Ancestors(最近公共祖先 LCA)

    POJ 1470 Closest Common Ancestors(最近公共祖先 LCA) Description Write a program that takes as input a root ...

  4. POJ 1470 Closest Common Ancestors (LCA,离线Tarjan算法)

    Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 13372   Accept ...

  5. POJ 1470 Closest Common Ancestors (LCA, dfs+ST在线算法)

    Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 13370   Accept ...

  6. POJ 1470 Closest Common Ancestors 【LCA】

    任意门:http://poj.org/problem?id=1470 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000 ...

  7. poj1470 Closest Common Ancestors [ 离线LCA tarjan ]

    传送门 Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 14915   Ac ...

  8. poj——1470 Closest Common Ancestors

    Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissions: 20804   Accept ...

  9. Closest Common Ancestors POJ 1470

    Language: Default Closest Common Ancestors Time Limit: 2000MS   Memory Limit: 10000K Total Submissio ...

随机推荐

  1. 51nod 1134最长递增子序列

    1134 最长递增子序列 基准时间限制:1 秒 空间限制:131072 KB 分值: 0 难度:基础题  收藏  关注 给出长度为N的数组,找出这个数组的最长递增子序列.(递增子序列是指,子序列的元素 ...

  2. 贪心/数学 Codeforces Round #212 (Div. 2) A. Two Semiknights Meet

    题目传送门 /* 贪心/数学:还以为是BFS,其实x1 + 4 * k = x2, y1 + 4 * l = y2 */ #include <cstdio> #include <al ...

  3. 转 PHP Cookies

    cookie 常用于识别用户. 什么是 Cookie? cookie 常用于识别用户.cookie 是服务器留在用户计算机中的小文件.每当相同的计算机通过浏览器请求页面时,它同时会发送 cookie. ...

  4. 440 K-th Smallest in Lexicographical Order 字典序的第K小数字

    给定整数 n 和 k,找到 1 到 n 中字典序第 k 小的数字.注意:1 ≤ k ≤ n ≤ 109.示例 :输入:n: 13   k: 2输出:10解释:字典序的排列是 [1, 10, 11, 1 ...

  5. 理解http浏览器的协商缓存和强制缓存

    阅读目录 一:浏览器缓存的作用是什么? 二:理解协商缓存 1 Last-Modified/if-Modify-Since 2 ETag/if-None-Match 三:理解强制缓存 回到顶部 一:浏览 ...

  6. .net 发送邮件验证码

    using System;using System.Collections.Generic;using System.Linq;using System.Net;using System.Net.Ma ...

  7. LoadRunner脚本回放与设置

    一.runtime setting 1.迭代次数设置与迭代步长(循环间隔时间) 2.日志打印设置       二.实时观看回放 1.动态回放与静态回放(静态回放时,不会有逐行高亮显示:动态回放时高亮显 ...

  8. php防止页面刷新代码

    //代理IP直接退出 empty($_SERVER['HTTP_VIA']) or exit('Access Denied'); //防止快速刷新 session_start(); $seconds ...

  9. CREATE TABLE - 定义一个新表

    SYNOPSIS CREATE [ [ GLOBAL | LOCAL ] { TEMPORARY | TEMP } ] TABLE table_name ( { column_name data_ty ...

  10. 1434:【例题2】Best Cow Fences

    1434:[例题2]Best Cow Fences 时间限制: 1000 ms         内存限制: 65536 KB提交数: 263     通过数: 146 [题目描述] 给定一个长度为n的 ...