HDU 3407.Zjnu Stadium 加权并查集
Zjnu Stadium
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3726 Accepted Submission(s): 1415
These days, Busoniya want to hold a large-scale theatrical performance in this stadium. There will be N people go there numbered 1--N. Busoniya has Reserved several seats. To make it funny, he makes M requests for these seats: A B X, which means people numbered B must seat clockwise X distance from people numbered A. For example: A is in column 4th and X is 2, then B must in column 6th (6=4+2).
Now your task is to judge weather the request is correct or not. The rule of your judgement is easy: when a new request has conflicts against the foregoing ones then we define it as incorrect, otherwise it is correct. Please find out all the incorrect requests and count them as R.
For every case:
The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.
Then M lines follow, each line has 3 integer A(1<=A<=N), B(1<=B<=N), X(0<=X<300) (A!=B), separated by a space.
Output R, represents the number of incorrect request.
Hint:
(PS: the 5th and 10th requests are incorrect)
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<vector>
#include<set>
using namespace std;
#define PI acos(-1.0)
typedef long long ll;
typedef pair<int,int> P;
const int maxn=1e5+,maxm=1e5+,inf=0x3f3f3f3f,mod=1e9+;
const ll INF=1e13+;
int pa[maxn];
int dist[maxn];
void init(int n)
{
for(int i=; i<=n; i++)
pa[i]=i,dist[i]=;
}
int findset(int x)
{
if(pa[x]==x) return x;
int fax=pa[x];
pa[x]=findset(pa[x]);
dist[x]+=dist[fax];
return pa[x];
}
void unit(int x,int y,int w)
{
int fx=findset(x),fy=findset(y);
pa[fx]=fy;
dist[fx]=dist[y]+w-dist[x];
}
bool same(int x,int y,int w)
{
if(findset(x)!=findset(y)) return false;
else if(dist[findset(x)]==dist[y]+w-dist[x]) return false;
else return true;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int ans=;
init(n);
for(int i=; i<=m; i++)
{
int a,b,x;
scanf("%d%d%d",&a,&b,&x);
if(same(a,b,x)) ans++;
else unit(a,b,x);
}
cout<<ans<<endl;
}
return ;
}
加权并查集
HDU 3407.Zjnu Stadium 加权并查集的更多相关文章
- hdu 3047 Zjnu Stadium(并查集)
题意: 300个座位构成一个圈. 有N个人要入座. 共有M个说明 :A B X ,代表B坐在A顺时针方向第X个座位上.如果这个说明和之前的起冲突,则它是无效的. 问总共有多少个无效的. 思路: 并查集 ...
- hdu3047 Zjnu Stadium (并查集)
Zjnu Stadium Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Tota ...
- hdu 3047 Zjnu Stadium(加权并查集)2009 Multi-University Training Contest 14
题意: 有一个运动场,运动场的坐席是环形的,有1~300共300列座位,每列按有无限个座位计算T_T. 输入: 有多组输入样例,每组样例首行包含两个正整数n, m.分别表示共有n个人,m次操作. 接下 ...
- Zjnu Stadium(加权并查集)
Zjnu Stadium Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Tota ...
- Hdu 2047 Zjnu Stadium(带权并查集)
Zjnu Stadium Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total ...
- hdu 3635 Dragon Balls(加权并查集)2010 ACM-ICPC Multi-University Training Contest(19)
这道题说,在很久很久以前,有一个故事.故事的名字叫龙珠.后来,龙珠不知道出了什么问题,从7个变成了n个. 在悟空所在的国家里有n个城市,每个城市有1个龙珠,第i个城市有第i个龙珠. 然后,每经过一段时 ...
- A Bug's Life(加权并查集)
Description Background Professor Hopper is researching the sexual behavior of a rare species of bug ...
- A Bug's Life(加权并查集)
Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs ...
- P1196 银河英雄传说(加权并查集)
P1196 银河英雄传说 题目描述 公元五八○一年,地球居民迁移至金牛座α第二行星,在那里发表银河联邦 创立宣言,同年改元为宇宙历元年,并开始向银河系深处拓展. 宇宙历七九九年,银河系的两大军事集团在 ...
随机推荐
- yii gii配置ip限制使用gii
<?php $config = [ 'components' => [ 'request' => [ // !!! insert a secret key in the follow ...
- HTML5将<video>视频设置为页面动态背景
<!DOCTYPE html><html><head> <title>Operation Aborted Example</title> & ...
- pandas 中的常用数学计算
1.开方: >>> s = pd.Series([1.2 + 1j]) >>> s.abs()
- msf客户端渗透(十):社会工程学
启动社会工程学攻击组件 生成二维码攻击模块 输入你想生成二维码的url,这里做演示用www.baidu.com 二维码生成后,在这个路径下 生成u盘,DVD的多媒体攻击载荷 通过修改autorun.i ...
- uniquefu Python+Selenium学习--select
场景 在处理下拉框(select)的时候selenium给我们提供了一系列的便捷方法,我们只需要使用selenium.webdriver.support.select.Select类来稍微封装一下就好 ...
- 【资料整理】ADO.NET
ADO.NET: 1.SYSTEM.DATA 命名空间下类的集合的统称,用于操作 连接数据库的 它提供了统一的编程接口,可以操作不同的数据库 数据库实例:默认实例(MSSQLSEVER)和命名实例(M ...
- http://www.bugku.com:Bugku——SQL注入1(http://103.238.227.13:10087/)
Bugku——SQL注入1(http://103.238.227.13:10087/) 过滤了几乎所有的关键字,尝试绕过无果之后发现,下面有个xss过滤代码.经搜索得该函数会去掉所有的html标签,所 ...
- 【go语言实现服务器接收http请求以及出现泄漏时的解决方案】
一.关于基础的程序的实现 刚开始的时候程序是这样实现的: // Hello package main import ( "database/sql" "fmt" ...
- TZOJ 3030 Courses(二分图匹配)
描述 Consider a group of N students and P courses. Each student visits zero, one or more than one cour ...
- UVa 1103 Ancient Messages(二重深搜)
In order to understand early civilizations, archaeologists often study texts written in ancient lang ...