链接:https://codeforces.com/contest/1167/problem/C

题意:

In some social network, there are nn users communicating with each other in mm groups of friends. Let's analyze the process of distributing some news between users.

Initially, some user xx receives the news from some source. Then he or she sends the news to his or her friends (two users are friends if there is at least one group such that both of them belong to this group). Friends continue sending the news to their friends, and so on. The process ends when there is no pair of friends such that one of them knows the news, and another one doesn't know.

For each user xx you have to determine what is the number of users that will know the news if initially only user xx starts distributing it.

思路:

裸并查集

代码:

#include <bits/stdc++.h>
using namespace std; const int MAXN = 5e5+10; int Fa[MAXN];
int Dis[MAXN]; int GetF(int x)
{
if (Fa[x] == x)
return Fa[x];
Fa[x] = GetF(Fa[x]);
return Fa[x];
} int main()
{
int n, m;
cin >> n >> m;
for (int i = 1;i <= n;i++)
Fa[i] = i;
int num, p;
for (int i = 1;i <= m;i++)
{
cin >> num;
int head;
if (num > 0)
{
cin >> p;
head = GetF(p);
num--;
}
while (num--)
{
cin >> p;
int h = GetF(p);
Fa[h] = head;
}
}
for (int i = 1;i <= n;i++)
{
int head = GetF(i);
Dis[head]++;
}
for (int i = 1;i <= n;i++)
{
int head = GetF(i);
cout << Dis[head] << ' ' ;
}
cout << endl; return 0;
}

  

Educational Codeforces Round 65 (Rated for Div. 2) C. News Distribution的更多相关文章

  1. Educational Codeforces Round 65 (Rated for Div. 2)题解

    Educational Codeforces Round 65 (Rated for Div. 2)题解 题目链接 A. Telephone Number 水题,代码如下: Code #include ...

  2. Educational Codeforces Round 65 (Rated for Div. 2) D. Bicolored RBS

    链接:https://codeforces.com/contest/1167/problem/D 题意: A string is called bracket sequence if it does ...

  3. Educational Codeforces Round 65 (Rated for Div. 2) B. Lost Numbers

    链接:https://codeforces.com/contest/1167/problem/B 题意: This is an interactive problem. Remember to flu ...

  4. Educational Codeforces Round 65 (Rated for Div. 2) A. Telephone Number

    链接:https://codeforces.com/contest/1167/problem/A 题意: A telephone number is a sequence of exactly 11  ...

  5. Educational Codeforces Round 65 (Rated for Div. 2)B. Lost Numbers(交互)

    This is an interactive problem. Remember to flush your output while communicating with the testing p ...

  6. [ Educational Codeforces Round 65 (Rated for Div. 2)][二分]

    https://codeforc.es/contest/1167/problem/E E. Range Deleting time limit per test 2 seconds memory li ...

  7. Educational Codeforces Round 65 (Rated for Div. 2)

    A:签到. #include<bits/stdc++.h> using namespace std; #define ll long long #define inf 1000000010 ...

  8. Educational Codeforces Round 65 (Rated for Div. 2) E. Range Deleting(思维+coding)

    传送门 参考资料: [1]:https://blog.csdn.net/weixin_43262291/article/details/90271693 题意: 给你一个包含 n 个数的序列 a,并且 ...

  9. Educational Codeforces Round 65 (Rated for Div. 2)(ACD)B是交互题,不怎么会

    A. Telephone Number A telephone number is a sequence of exactly 11 digits, where the first digit is  ...

随机推荐

  1. matlab之结构体数组struct

    以下内容来自于:https://blog.csdn.net/u010999396/article/details/54413615/ 要在MALTAB中实现比较复杂的编程,就不能不用struct类型. ...

  2. Web性能测试中的几个概念【转】

    每秒查询率QPS:对一个特定的查询服务器在规定时间内所处理流量多少的衡量标准,即每秒请求数,即最大谈吐能力. 并发数:并发数和QPS是不同的概念,一般说QPS会说多少并发用户下QPS,当QPS相同时, ...

  3. Java中内部类中使用外面变量为什么final修饰?

    所以final从语法上约束了实际上两个不同变量的一致性(表现为同一变量). 我的理解: 例如这样的代码: public void test(String str){ str="abc&quo ...

  4. 1>/dev/null 2>&1 & 意思解析

    原文:https://jingyan.baidu.com/article/6dad5075334e26a123e36e31.html 用 /dev/null 2>&1 这样的写法.这条命 ...

  5. codeforces 609D D. Gadgets for dollars and pounds(二分+贪心)

    题目链接: D. Gadgets for dollars and pounds time limit per test 2 seconds memory limit per test 256 mega ...

  6. 机器学习:Colorization using Optimization

    今天介绍 Siggraph 2004 年的一篇文章: Colorization using Optimization,利用优化的方法对灰度图像进行着色,这里用到了非常经典的泊松方程以及稀疏矩阵的线性优 ...

  7. OpenCV——PS滤镜之 波浪效果 wave

    // define head function #ifndef PS_ALGORITHM_H_INCLUDED #define PS_ALGORITHM_H_INCLUDED #include < ...

  8. appium 特殊操作

      一.触摸操作   1.driver.tap([坐标],持续点击时间) 除了定位到元素的点击外,也可以通过tab实现坐标的点击 driver.tap(driver.tap([(216,1776)], ...

  9. ACM学习历程—BestCoder 2015百度之星资格赛1006 单调区间(组合数学)

    Problem Description 百小度最近在逛博客,然后发现了一个有趣的问题. 如下图所示,是一个12 位数014326951987 , 它的数字先逐渐变大, 然后变小,再变大,接着变小,又变 ...

  10. 【C++基础】浅拷贝和深拷贝

    简单理解: 对于一块内存,浅拷贝只是增加了一个指针,这样两个变量都指向这块内存,二深拷贝则是先开辟一块同等大小的新内存区,将待拷贝内存的内容复制过来,再赋予一个指向新内存的指针.区别在于:浅拷贝会造成 ...