Codeforces Gym 100342J Problem J. Triatrip 求三元环的数量 bitset
Problem J. Triatrip
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/gym/100342/attachments
Description
The travel agency “Four Russians” is offering the new service for their clients. Unlike other agencies that only suggest one-way or roundtrip for airline tickets to their customers, “Four Russians” offers the brand new idea — triatrip. Triatrip traveler starts in some city A, flies to some city B, then flies to some city C, and returns to the city A.
Now the managers of the agency started to wonder, how many different triatrips they can offer to their customers. Given a map of all possible flights, help them to find that out.
Input
The first line of the input file contains two integer numbers n — the number of cities that are served by airlines that agree to sell their tickets via the agency (3 ≤ n ≤ 1500). The following n lines contain a sequence of n characters each — the j-th character of the i-th line is ‘+’ if it is possible to fly from the i-th city to the j-th one, and ‘-’ if it is not. The i-th character of the i-th line is ‘-’.
Output
Output one integer number — the number of triatrips that the agency can offer to its customers.
Sample Input
4
--+-
+--+
-+--
--+-
Sample Output
2
HINT
题意
给出邻接矩阵,有向图,找出三元环的个数
题解:
bitset 暴力枚举一条边,出度入度的集合用到bitset
代码:
//作者:1085422276
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include<bits/stdc++.h>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
const int inf = ;
inline ll read()
{
ll x=,f=;
char ch=getchar();
while(ch<''||ch>'')
{
if(ch=='-')f=-;
ch=getchar();
}
while(ch>=''&&ch<='')
{
x=x*+ch-'';
ch=getchar();
}
return x*f;
}
ll exgcd(ll a,ll b,ll &x,ll &y)
{
ll temp,p;
if(b==){
x=;y=;
return a;
}
p=exgcd(b,a%b,x,y);
temp=x;x=y;y=temp-(a/b)*y;
return p;
}
//*******************************
#define N 1510
bitset<N> in[N],out[N],he;
char mp[][];
int main()
{ freopen("triatrip.in","r",stdin);
freopen("triatrip.out","w",stdout);
int n;
n=read();
for(int i=;i<=n;i++)
scanf("%s",mp[i]);
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
if(i==j)continue;
if(mp[i][j-]=='+')
{
in[i][j]=true;
out[j][i]=true;
}
}
}
ll ans=;
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
if(i==j)continue;
if(mp[i][j-]=='+')
he=(in[j])&(out[i]),ans+=he.count();
}
}
cout<<ans/<<endl;
return ;
}
Codeforces Gym 100342J Problem J. Triatrip 求三元环的数量 bitset的更多相关文章
- Codeforces Gym 100342J Problem J. Triatrip bitset 求三元环的数量
Problem J. TriatripTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100342/att ...
- Codeforces Gym 100342J Problem J. Triatrip 三元环
题目链接: http://codeforces.com/gym/100342 题意: 求三元环的个数 题解: 用bitset分别统计每个点的出度的边和入度的边. 枚举每一条边(a,b),计算以b为出度 ...
- Gym 100342J Triatrip (求三元环的数量) (bitset优化)
<题目链接> 题目大意:用用邻接矩阵表示一个有向图,现在让你求其中三元环的数量. 解题分析:先预处理得到所有能够直接到达每个点的集合$arrive[N]$和所有能够由当前点到达的集合$to ...
- Gym - 100342J:Triatrip(Bitset加速求三元环的数量)
题意:求有向图里面有多少个三元环. 思路:枚举起点A,遍历A可以到的B,然后求C的数量,C的数量位B可以到是地方X集合,和可以到A的地方Y集合的交集(X&Y). B点可以枚举,也可以遍历.(两 ...
- Gym - 100342J Triatrip (bitset求三元环个数)
https://vjudge.net/problem/Gym-100342J 题意:给出一个邻接矩阵有向图,求图中的三元环的个数. 思路: 利用bitset暴力求解,记得最后需要/3. #includ ...
- Codeforces 434E - Furukawa Nagisa's Tree(三元环+点分治)
Codeforces 题面传送门 & 洛谷题面传送门 场号 hopping,刚好是我的学号(指 round 的编号) 注:下文中分别用 \(X,Y,K\) 代替题目中的 \(x,y,k\) 注 ...
- Codeforces Gym 100342C Problem C. Painting Cottages 转化题意
Problem C. Painting CottagesTime Limit: 2 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/10 ...
- Codeforces Gym 100342C Problem C. Painting Cottages 暴力
Problem C. Painting CottagesTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/1 ...
- Codeforces Gym 100610 Problem A. Alien Communication Masterclass 构造
Problem A. Alien Communication Masterclass Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codefo ...
随机推荐
- ccleaner注册码
名称:Registered User 密钥:CBB4-FJN4-EPC6-G5P6-QT4C 先不要启动程序,然后断网注册.
- block与函数指针有什么区别
block就是一个代码块,但是它的神奇之处在于在内联(inline)执行的时候(这和C++很像)还可以传递参数.同时block本身也可以被作为参数在方法和函数间传递,这就给予了block无限的可能. ...
- C#实现AES加解密方法
using System; using System.Collections.Generic; using System.Text; using System.Security.Cryptograph ...
- ftp的20 21端口和主动被动模式
ftp只支持tcp连接,不支持udp连接. ftp使用两个端口: 21(控制端口, 命令端口) , 20(数据端口) 21端口: 用来控制用户验证, 连接的建立和关闭:open/close/bye ...
- python闭包小例子
------------------ 首先根据实例, 体会一下闭包的效果 ------------------ 定义闭包: def foo(x): a = [0] def bar(y): a[0] = ...
- Flume-NG内置计数器(监控)源码级分析
Flume的内置监控怎么整?这个问题有很多人问.目前了解到的信息是可以使用Cloudera Manager.Ganglia有图形的监控工具,以及从浏览器获取json串,或者自定义向其他监控系统汇报信息 ...
- BC.36.Gunner(hash)
Gunner Accepts: 391 Submissions: 1397 Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536 ...
- why we use Symbols in Hash
Rather than using Strings as the keys in a Hash, it’s better practice to use Symbols. Symbols are ju ...
- :( Call to a member function Table() on a non-object 错误位置
:( Call to a member function Table() on a non-object 错误位置 $Model不是模板,是你自己先前链接数据库返回的对象...我的是改为$Form
- 从零开始写一个武侠冒险游戏-7-用GPU提升性能(2)
从零开始写一个武侠冒险游戏-7-用GPU提升性能(2) ----把地图处理放在GPU上 作者:FreeBlues 修订记录 2016.06.21 初稿完成. 2016.08.06 增加对 XCode ...