HDU3047 Zjnu Stadium


Problem 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:

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


就是有一个环,给出多组关系满足a=b+x,判断不满足的有多少组

然后带权并查集维护一下,主义在合并两个并查集的时候差量的计算


 #include<bits/stdc++.h>
using namespace std;
#define N 50010
int fa[N],dis[N];
int n,m;
int Find(int x){
if(fa[x]==x)return x;
int f=Find(fa[x]);
dis[x]+=dis[fa[x]];
return fa[x]=f;
}
void init(){for(int i=;i<=n;i++)fa[i]=i,dis[i]=;}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
init();
int res=;
while(m--){
int a,b,x;
scanf("%d%d%d",&a,&b,&x);
int f_a=Find(a);
int f_b=Find(b);
if(f_a==f_b){
if(dis[a]+x!=dis[b])res++;
continue;
}
dis[f_b]=dis[a]-dis[b]+x;
fa[f_b]=f_a;
}
printf("%d\n",res);
}
return ;
}

HDU3047 Zjnu Stadium 【带权并查集】的更多相关文章

  1. HDU3047 Zjnu Stadium 带权并查集

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

  2. 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 ...

  3. 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 ...

  4. 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. ...

  5. Zjnu Stadium(hdu3047带权并查集)

    题意:一个300列的无限行的循环场地,a b d代表a,b顺时针相距d的距离,现在给你一些距离,判断是否有冲突,如果有冲突计算冲突的次数 思路:带权并查集 a,b的距离等于b到根节点的距离 - a到根 ...

  6. HDU 3047 Zjnu Stadium(带权并查集)

    http://acm.hdu.edu.cn/showproblem.php?pid=3047 题意: 给出n个座位,有m次询问,每次a,b,d表示b要在a右边d个位置处,问有几个询问是错误的. 思路: ...

  7. 【带权并查集】HDU 3047 Zjnu Stadium

    http://acm.hdu.edu.cn/showproblem.php?pid=3047 [题意] http://blog.csdn.net/hj1107402232/article/detail ...

  8. 【HDOJ3047】Zjnu Stadium(带权并查集)

    题意:浙江省第十二届大学生运动会在浙江师范大学举行,为此在浙师大建造了一座能容纳近万人的新体育场. 观众席每一行构成一个圆形,每个圆形由300个座位组成,对300个座位按照顺时针编号1到300,且可以 ...

  9. HDU 3047 带权并查集 入门题

    Zjnu Stadium 题目链接 http://acm.hdu.edu.cn/showproblem.php?pid=3047 Problem Description In 12th Zhejian ...

随机推荐

  1. linux一键安装nginx脚本

    #!/bin/sh echo "----------------------------------start install nginx ------------------------- ...

  2. 在Jupyter notebook中使用特定虚拟环境中的python的kernel

        在虚拟环境tf中安装完tensorflow后,在虚拟环境tf打开的jupyter里发现只有一个kernel-python3,新建一个文件, import tensorflow as tf ,发 ...

  3. 毕业设计总结(1)-canvas画图

    去年6月底完成的毕业设计,到现在也才开始给它做个总结,里面有很多可以学习和借鉴的东西. 我的毕业设计的题目是“一种路径规划算法的改进与设计”,具体的要求可参见下面的表格: 题目 一种路径规划算法的改进 ...

  4. import sys sys.path.append(...)

    模块搜索路径: 当我们试图加载一个模块时,Python会在指定的路径下搜索对应的.py文件,如果找不到,就会报错 默认情况下,Python解释器会搜索当前目录.所有已安装的内置模块和第三方模块,搜索路 ...

  5. Python 实现99乘法表

    首先,我们来回忆一下99乘法表长什么样子吧 进入正题:实现99乘法表 一.For循环 for i in range(1,10): for j in range(1,i+1): print(" ...

  6. 二十八 Python分布式爬虫打造搜索引擎Scrapy精讲—cookie禁用、自动限速、自定义spider的settings,对抗反爬机制

    cookie禁用 就是在Scrapy的配置文件settings.py里禁用掉cookie禁用,可以防止被通过cookie禁用识别到是爬虫,注意,只适用于不需要登录的网页,cookie禁用后是无法登录的 ...

  7. HDU 5289 尺取

    Assignment Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total ...

  8. STL学习笔记(转,还是比较全的)

    STL简介 1 概况 2 1.1 STL是什么 2 1.2 为什么我们需要学习STL 2 1.3 初识STL 2 1.4 STL 的组成 5 2 容器 6 2.1 基本容器——向量(vector) 6 ...

  9. Django 中设置分页页码,只显示当前页以及左右两页

    设置后的效果如下: Django 给我们提供了分页的功能:`Paginator`和`Page`类都是用来做分页的.他们在Django中的路径为:`from django.core.paginator ...

  10. 013PHP基础知识——流程控制(一)

    <?php /** * 13 流程控制(一) * if语句: if(表达式){ 表达式 }elseif(表达式){ 代码段 } * if语句中,一个条件成立,其他分支不执行. * if中的表达式 ...