HDOJ 题目3047 Zjnu Stadium（带权值并查集）

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1733    Accepted Submission(s): 655

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

   
    
  

 

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

 

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题意：

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。

ac代码

#include<stdio.h>
#include<string.h>
int pre[50005],d[50005];//d数组表示的是点到根节点的距离
int find(int x)
{
	int m;
	if(pre[x]==x)
		return x;
	m=pre[x];
	pre[x]=find(pre[x]);
	d[x]+=d[m];//f[4]=3,f[3]=1,f[1]=1;d[4]=d[4]+d[3]+d[1];  
	return pre[x];
}
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int a,b,c,i,j,ans=0;
		for(i=0;i<=n;i++)
		{
			pre[i]=i;
			d[i]=0;
		}
		while(m--)
		{
			int fa,fb;
			scanf("%d%d%d",&a,&b,&c);
			fa=find(a);
			fb=find(b);
			if(fa!=fb)
			{
				pre[fb]=fa; // 1 2 150,d[2]=150;3 4 200,d[4]=200;2 4 250,d[3]=200  
				d[fb]=c-d[b]+d[a];  // 1--150--2--50--3--200--4;d[3]=250-d[4]+d[2];  
			}
			else
			{
				if(d[b]-d[a]!=c)
					ans++;
			}
		}
		printf("%d\n",ans);
	}
}