CodeForces - 154C:Double Profiles (hash+排序)
You have been offered a job in a company developing a large social network. Your first task is connected with searching profiles that most probably belong to the same user.
The social network contains n registered profiles, numbered from 1 to n. Some pairs there are friends (the "friendship" relationship is mutual, that is, if i is friends with j, then j is also friends with i). Let's say that profiles i and j (i ≠ j) are doubles, if for any profile k (k ≠ i, k ≠ j) one of the two statements is true: either k is friends with i and j, or k isn't friends with either of them. Also, i and j can be friends or not be friends.
Your task is to count the number of different unordered pairs (i, j), such that the profiles i and j are doubles. Note that the pairs are unordered, that is, pairs (a, b) and (b, a) are considered identical.
Input
The first line contains two space-separated integers n and m (1 ≤ n ≤ 106, 0 ≤ m ≤ 106), — the number of profiles and the number of pairs of friends, correspondingly.
Next m lines contains descriptions of pairs of friends in the format "v u", where v and u (1 ≤ v, u ≤ n, v ≠ u) are numbers of profiles that are friends with each other. It is guaranteed that each unordered pair of friends occurs no more than once and no profile is friends with itself.
Output
Print the single integer — the number of unordered pairs of profiles that are doubles.
Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use the %I64d specificator.
Examples
3 3
1 2
2 3
1 3
3
3 0
3
4 1
1 3
2
Note
In the first and second sample any two profiles are doubles.
In the third sample the doubles are pairs of profiles (1, 3) and (2, 4).
题意:给定无向图,问有多少对点对,在不考虑它们之间的连边情况下,满足它们的直接连边的情况相同。
思路:用hash保存每个点的连边情况,然后排序得到hash值,相同的一并处理。
有两类:第一类是点对间有边,那么我们需要向自己加一条边,那么它们的连边情况是相同的。
第二类是点对间无边,直接处理即可。
#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define uint unsigned int
using namespace std;
const int maxn=;
const int seed=;
const int seed2=;
uint p[maxn],p2[maxn];long long ans;
pair<uint,uint>h[maxn];
int main()
{
int N,M,u,v;
scanf("%d%d",&N,&M); p[]=; p2[]=;
rep(i,,N) p[i]=p[i-]*seed;
rep(i,,N) p2[i]=p2[i-]*seed2;
rep(i,,N) h[i].first=p[i],h[i].second=p2[i];
rep(i,,M){
scanf("%d%d",&u,&v);
h[u].first+=p[v]; h[v].first+=p[u];
h[u+N].first+=p[v]; h[v+N].first+=p[u];
h[u].second+=p2[v]; h[v].second+=p2[u];
h[u+N].second+=p2[v]; h[v+N].second+=p2[u];
}
sort(h+,h+N+N+);
for(int i=;i<=N+N;i++){
int j=i;
while(j+<=N+N&&h[j+]==h[i]) j++;
ans+=(long long)(j-i+)*(j-i)/;
i=j;
}
printf("%I64d\n",ans);
return ;
}
CodeForces - 154C:Double Profiles (hash+排序)的更多相关文章
- Codeforces Round #109 (Div. 2) E. Double Profiles hash
题目链接: http://codeforces.com/problemset/problem/155/E E. Double Profiles time limit per test 3 second ...
- xtu summer individual 2 E - Double Profiles
Double Profiles Time Limit: 3000ms Memory Limit: 262144KB This problem will be judged on CodeForces. ...
- codeforces Round #440 A Search for Pretty Integers【hash/排序】
A. Search for Pretty Integers [题目链接]:http://codeforces.com/contest/872/problem/A time limit per test ...
- sort 树 hash 排序
STL 中 sort 函数用法简介 做 ACM 题的时候,排序是一种经常要用到的操作.如果每次都自己写个冒泡之类的 O(n^2) 排序,不但程序容易超时,而且浪费宝贵的比赛时间,还很有可能写错. ST ...
- [置顶] Codeforces 70D 动态凸包 (极角排序 or 水平序)
题目链接:http://codeforces.com/problemset/problem/70/D 本题关键:在log(n)的复杂度内判断点在凸包 或 把点插入凸包 判断:平衡树log(n)内选出点 ...
- DataTable 中varchar 转换为 Double 后重新 排序。
DataTable 查询出某个字段为varchar 类型的.不过里面存的为数字,需要进行排序.可是如果直接排序就会不对.因为为varchar类型的,需要转换一下. 方法一: dt.Columns.A ...
- ruby中Hash排序
当values都是整形时,按照Hash的Values排序: h = {'a'=>1,'b'=>2,'c'=>5,'d'=>4} h.sort {|a,b| a[1]<=& ...
- Codeforces 484B Maximum Value(排序+二分)
题目链接: http://codeforces.com/problemset/problem/484/B 题意: 求a[i]%a[j] (a[i]>a[j])的余数的最大值 分析: 要求余数的最 ...
- Codeforces 825E Minimal Labels - 拓扑排序 - 贪心
You are given a directed acyclic graph with n vertices and m edges. There are no self-loops or multi ...
随机推荐
- redis的安装配置(转)
一.下载windows版本的Redis 去官网找了很久,发现原来在官网上可以下载的windows版本的,现在官网以及没有下载地址,只能在github上下载,官网只提供linux版本的下载 官网下载地址 ...
- android开发之网络棋牌类在线游戏开发心得(服务器端、Java) 好文章值得收藏
标签: android服务器 2013-10-09 17:28 3618人阅读 评论(0) 收藏 举报 分类: android(11) 转自:http://blog.csdn.net/bromon/a ...
- ODS
一般在带有ODS的系统体系结构中,ODS都设计为如下几个作用: 1.在业务系统和数据仓库之间形成一个隔离层 一般的数据仓库应用系统都具有非常复杂的数据来源,这些数据存放在不同的地理位置.不同的数据库. ...
- springboot-controller的使用
获取url中的数据: @RestController public class HelloController { @RequestMapping(value="/say/{id}" ...
- 混淆Android JAR包的方法
http://blog.csdn.net/vurtne_ye/article/details/35599491 1)导出jar包 如何在eclipse上Android工程中导出jar包?google, ...
- HAproxy 介绍
HAproxy 介绍 (1)HAProxy 是一款提供高可用性.负载均衡以及基于TCP(第四层)和HTTP(第七层)应用的代理软件,支持虚拟主机,它是免费.快速并且可靠的一种解决方案. HAProxy ...
- java 跨数据库导入大数据
java 跨数据库导入大数据 /** * java程序跨服务器跨数据库批量导入导出百万级数据 * @param args * @throws Exception */ public static vo ...
- 关于series的统计
1.统计se中各个元素出现的次数: se.value_counts() 这个命令就非常重要!(df没有这个命令) 另外series转dict是to_dict(),注意与数组转list的tolist() ...
- b树的实现(c++)
转自:http://blog.chinaunix.net/uid-20196318-id-3030529.html B树的定义 假设B树的度为t(t>=2),则B树满足如下要求:(参考算法导论) ...
- JDK与JRE及其在Eclipse中的使用
转载自:http://blog.csdn.net/gx1058742912/article/details/51033942 JDK与jRE的区别 JDK(java development kit): ...