M - Zjnu Stadium

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

In 12th Zhejiang College Students Games 2007, there was a new stadium built in Zhejiang Normal University. It was a modern stadium which could hold thousands of people. The audience Seats made a circle. The total number of columns were 300 numbered 1--300, counted clockwise, we assume the number of rows were infinite.
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.
 

Input

There are many test cases:

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

For every case:

Output R, represents the number of incorrect request.
 

Sample Input

10 10
1 2 150
3 4 200
1 5 270
2 6 200
6 5 80
4 7 150
8 9 100
4 8 50
1 7 100
9 2 100
 

Sample Output

2

Hint

 Hint: (PS: the 5th and 10th requests are incorrect) 

一个类似的题:
https://www.cnblogs.com/yinbiao/p/9460772.html code:
#include<queue>
#include<set>
#include<cstdio>
#include <iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<string>
#include<string.h>
#include<memory>
using namespace std;
#define max_v 50005
#define INF 9999999
int pa[max_v];
int sum[max_v];
int n,m;
int ans;
void init()
{
for(int i=;i<=n;i++)
{
pa[i]=i;
sum[i]=;
}
}
int find_set(int x)
{
if(pa[x]!=x)
{
int t=pa[x];
pa[x]=find_set(pa[x]);
sum[x]+=sum[t];//!!!
}
return pa[x];
}
void union_set(int a,int b,int v)
{
int x=find_set(a);
int y=find_set(b);
if(x==y)
{
if(sum[a]-sum[b]!=v)//!!!
ans++;
}else
{
pa[x]=y;
sum[x]=sum[b]-sum[a]+v;//!!!
}
}
int main()
{
while(~scanf("%d %d",&n,&m))
{
int x,y,w;
ans=;
init();
for(int i=;i<m;i++)
{
scanf("%d %d %d",&x,&y,&w);
union_set(x,y,w);
}
printf("%d\n",ans);
}
return ;
}

HDU 3047 Zjnu Stadium(带权并查集,难想到)的更多相关文章

  1. hdu 3047–Zjnu Stadium(带权并查集)

    题目大意: 有n个人坐在zjnu体育馆里面,然后给出m个他们之间的距离, A B X, 代表B的座位比A多X. 然后求出这m个关系之间有多少个错误,所谓错误就是当前这个关系与之前的有冲突. 分析: 首 ...

  2. HDU 3047 Zjnu Stadium(带权并查集)

    题意:有一个环形体育场,有n个人坐,给出m个位置关系,A B x表示B所在的列在A的顺时针方向的第x个,在哪一行无所谓,因为假设行有无穷个. 给出的座位安排中可能有与前面矛盾的,求有矛盾冲突的个数. ...

  3. 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 ...

  4. 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 ...

  5. HDU3047 Zjnu Stadium 带权并查集

    转:http://blog.csdn.net/shuangde800/article/details/7983965 #include <cstdio> #include <cstr ...

  6. hdu 5441 Travel 离线带权并查集

    Travel Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5441 De ...

  7. How Many Answers Are Wrong (HDU - 3038)(带权并查集)

    题目链接 并查集是用来对集合合并查询的一种数据结构,或者判断是不是一个集合,本题是给你一系列区间和,判断给出的区间中有几个是不合法的. 思考: 1.如何建立区间之间的联系 2.如何发现悖论 首先是如何 ...

  8. hdu 5441 travel 离线+带权并查集

    Time Limit: 1500/1000 MS (Java/Others)  Memory Limit: 131072/131072 K (Java/Others) Problem Descript ...

  9. hdu 2818 Building Block (带权并查集,很优美的题目)

    Problem Description John are playing with blocks. There are N blocks ( <= N <= ) numbered ...N ...

  10. hdu 3635 Dragon Balls (带权并查集)

    Dragon Balls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tota ...

随机推荐

  1. C/C++:判断机器是32位还是64位

    要求是不使用sizeof,一开始写了个看似可以,但是有问题的方法: long* a = NULL; ; int n = (char*)b - (char*)a; 这个方法等价于sizeof(long) ...

  2. springcloud 集成kafka问题记录,发消息报错:ERROR o.s.kafka.support.LoggingProducerListener - Exception thrown when sending a message with key='null' and payload='{-1,

    在springcloud集成kafka,发送消息时报错: 2018-08-15 16:01:34.159 [http-nio-8081-exec-1] INFO  org.apache.kafka.c ...

  3. H5实现拍照上传功能

    <input type="file" capture="camera" accept="image/*" >

  4. windows10(本机)与VirtualBox中CentOS7(虚拟机)互相访问总结

    先把我这里的环境说下: 本机(windows10),发布了一个tomcat服务:http://192.168.0.106:8080/axis/services/VPMService?wsdl 如下图: ...

  5. Apache 2 解析html中的php

    Ubuntu下安装Apache 2无法解析html中的php Ubuntu下安装了Apache 2却无法解析html中的php ,好多说是在httpd.conf文件中修改代码,但是ubuntu中没有这 ...

  6. Install Python on Mac

    1. 从官网下载最新版Python 3.X 后安装:由于Mac OS X EI Capitan中默认已经集成了 Python 2.7,因此需要在Terminal中输入 Python3 来检测是否安装成 ...

  7. [转]Linux芯片级移植与底层驱动(基于3.7.4内核)

      1.   SoC Linux底层驱动的组成和现状 为了让Linux在一个全新的ARM SoC上运行,需要提供大量的底层支撑,如定时器节拍.中断控制器.SMP启动.CPU hotplug以及底层的G ...

  8. 移动端App开发 - 01 - 开篇

    移动端App开发 - 01 - 开篇 从此笔记之后开启移动端 app 开发学习 该系列笔记去掉所有无关重要的东西,简介干练 我的移动端App开发笔记 1.移动端App开发 - 02 - iPhone/ ...

  9. moveTaskToback退后台的用法及作用

    1 方法:public boolean moveTaskToBack(boolean nonRoot) activity里有这个方法,参数说明如下: nonRoot=false→ 仅当activity ...

  10. php 空格无法替换,utf-8空格惹的祸

    一次坑爹的小bug.读取一段文字(编码utf-8),想替换掉空格,str_replace(" "..).preg_replace("/\s/"..)都不起作用. ...