G. Suggested Friends
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarpus works as a programmer in a start-up social network. His boss gave his a task to develop a mechanism for determining suggested friends. Polycarpus thought much about the task and came to the folowing conclusion.

Let's say that all friendship relationships in a social network are given as m username pairs ai, bi (ai ≠ bi). Each pair ai, bi means that users ai and bi are friends. Friendship is symmetric, that is, if ai is friends with bi, then bi is also friends with ai. User y is a suggested friend for user x, if the following conditions are met:

  1. x ≠ y;
  2. x and y aren't friends;
  3. among all network users who meet the first two conditions, user y has most of all common friends with user x. User z is a common friend of user x and user y (z ≠ x, z ≠ y), if x and z are friends, and y and z are also friends.

Your task is to help Polycarpus to implement a mechanism for determining suggested friends.

Input

The first line contains a single integer m (1 ≤ m ≤ 5000) — the number of pairs of friends in the social network. Next m lines contain pairs of names of the users who are friends with each other. The i-th line contains two space-separated names ai and bi (ai ≠ bi). The users' names are non-empty and consist of at most 20 uppercase and lowercase English letters.

It is guaranteed that each pair of friends occurs only once in the input. For example, the input can't contain xy and yx at the same time. It is guaranteed that distinct users have distinct names. It is guaranteed that each social network user has at least one friend. The last thing guarantees that each username occurs at least once in the input.

Output

In the first line print a single integer n — the number of network users. In next n lines print the number of suggested friends for each user. In the i-th line print the name of the user ci and the number of his suggested friends di after a space.

You can print information about the users in any order.

题意

定义A是B的推荐朋友,当且仅当A拥有B最多的共同朋友,且A和B不是朋友。问每个人有多少推荐朋友

题解

由于边很少,只有5000条,那么可以暴力枚举两个点,再暴力判一下就好,这样绝对不会超时,至于为什么,大家可以脑补脑补。

代码

#include<iostream>
#include<bitset>
#include<vector>
#include<cstring>
#include<string>
#include<map>
#include<cstdio>
#define MAX_N 2*5555
using namespace std; map<string,int> ma; string name[MAX_N]; int n,m; bitset<MAX_N> bi[MAX_N];
vector<int> G[MAX_N]; char cc[]; int main() {
scanf("%d", &m);
for (int i = ; i < m; i++) {
string u, v;
scanf("%s", cc);
u = cc;
scanf("%s", cc);
v = cc;
if (ma.find(u) == ma.end())ma[u] = n++;
if (ma.find(v) == ma.end())ma[v] = n++;
int s = ma[u], t = ma[v];
name[s] = u;
name[t] = v;
bi[s][t] = bi[t][s] = ;
G[t].push_back(s);
G[s].push_back(t);
}
printf("%d\n", n);
for (int i = ; i < n; i++) {
printf("%s ", name[i].c_str());
int c = ;
int mf = ;
for (int j = ; j < n; j++) {
if (bi[i][j] || i == j)continue;
int tmp = ;
for(int k=;k<G[j].size();k++)
if(bi[i][G[j][k]])tmp++;
if (tmp > mf) {
mf = tmp;
c = ;
}
else if (tmp == mf)c++;
}
printf("%d\n", c);
}
return ;
}

Codeforces 245G Suggested Friends 暴力乱搞的更多相关文章

  1. VIJOS1476 旅行规划(树形Dp + DFS暴力乱搞)

    题意: 给出一个树,树上每一条边的边权为 1,求树上所有最长链的点集并. 细节: 可能存在多条最长链!最长链!最长链!重要的事情说三遍 分析: 方法round 1:暴力乱搞Q A Q,边权为正-> ...

  2. Codeforces 538G - Berserk Robot(乱搞)

    Codeforces 题目传送门 & 洛谷题目传送门 一道很神的乱搞题 %%% 首先注意到如果直接去做,横纵坐标有关联,不好搞.这里有一个非常套路的技巧--坐标轴旋转,我们不妨将整个坐标系旋转 ...

  3. Codeforces 306D - Polygon(随机化+乱搞)

    Codeforces 题目传送门 & 洛谷题目传送门 中考终于结束了--简单写道题恢复下状态罢. 首先这一类题目肯定没法用一般的方法解决,因此考虑用一些奇淫的乱搞做法解决这道题,不难发现,如果 ...

  4. codeforces 665C C. Simple Strings(乱搞)

    题目链接: C. Simple Strings time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

  5. [CSP-S模拟测试]:Cicada拿衣服(暴力+乱搞)

    题目传送门(内部题94) 输入格式 第一行两个整数$n,k$,代表衣服的数量和阈值. 接下来一行$n$个数,第$i$个数$a_i$表示每件衣服的愉悦值. 输出格式 输出一行$n$个数,第$i$个数为$ ...

  6. Codeforces #254 div1 B. DZY Loves FFT 暴力乱搞

    B. DZY Loves FFT 题目连接: http://codeforces.com/contest/444/problem/B Description DZY loves Fast Fourie ...

  7. Codeforces Gym 100203G Good elements 暴力乱搞

    原题链接:http://codeforces.com/gym/100203/attachments/download/1702/statements.pdf 题解 考虑暴力的复杂度是O(n^3),所以 ...

  8. Codeforces 34C-Page Numbers(set+vector+暴力乱搞)

    C. Page Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard inp ...

  9. BZOJ 1491: [NOI2007]社交网络(Floyd+暴力乱搞)

    题面: https://www.lydsy.com/JudgeOnline/problem.php?id=1491 题解: 先看数据范围,n<=100..欸可以乱搞了 首先因为小学学过的乘法原理 ...

随机推荐

  1. leetcode-4-basic

    解题思路:这道题比较简单,代码不贴了.需要注意的是: 数字与字符串之间的转换, char str[100]; sprintf(str, "%d", num); 解题思路: 这道题是 ...

  2. HDU:2255-奔小康赚大钱(KM算法模板)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=2255 奔小康赚大钱 Time Limit: 1000/1000 MS (Java/Others) Mem ...

  3. Linux学习-开放源码的软件安装与升级简介

    什么是开放源码.编译程序与可执行文件 我们说过,在 Linux 系统上面,一个文件能不能被执行看的是有没有可执行的那个权限 (具有 x permission),不过,Linux 系统上真 正认识的可执 ...

  4. CodeForces 14D 树的直径 Two Paths

    给出一棵树,找出两条不相交即没有公共点的路径,使得两个路径的长度的乘积最大. 思路:枚举树中的边,将该边去掉,分成两棵树,分别求出这两棵树的直径,乘起来维护一个最大值即可. #include < ...

  5. TextView设置缩略显示

    1.代码设置 textview.setSingleLine(); textview.setEllipsiz(TextUtils.TruncateAt.valueOf("END")) ...

  6. Python面试题(练习一)

    1.Python的可变类型和不可变类型? 可变类型:list.dict(列表和字典) 不可变类型:数字.字符串.元组 2.求结果: v = dict.fromkeys(['k1','k2'],[]) ...

  7. Mongodb 删除记录里的某个字段

    //例如要把User表中address字段删除 db.User.update({},{$unset:{'address':''}},false, true)

  8. 2017"百度之星"程序设计大赛 - 初赛(A)

    小C的倍数问题  Accepts: 1990  Submissions: 4931  Time Limit: 2000/1000 MS (Java/Others)  Memory Limit: 327 ...

  9. Model View Controller(MVC) in PHP

    The model view controller pattern is the most used pattern for today’s world web applications. It ha ...

  10. 浏览器提示ERR_CONTENT_DECODING_FAILED,Gzip压缩数据无法解压

    最近在页面上有个显示数据表格的功能,数据由后台传给前台JS表格插件.数据格式为JSON 由于数据量很大,就想到用GZIP压缩以后传给前台.压缩前,某个表格的数据量达到3M多,用GZIP压缩后就200K ...