【杭电oj】3047 - Zjnu Stadium（带权并查集）

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2464    Accepted Submission(s): 942

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

还是带权并查集，在合并的部分有时候还是分不太清，还需要多练习啊！

代码如下：

#include <cstdio>
#include <algorithm>
using namespace std;
int f[50011];
int sum[50011];		//与其根节点的距离 
int find(int x)
{
	if (x!=f[x])
	{
		int t=f[x];
		f[x]=find(f[x]);
		sum[x]+=sum[t];
	}
	return f[x];
}
int main()
{
	int n,m;		//座位数，语句条数
	int x,y,l;
	int fx,fy;
	int ans;
	while (~scanf ("%d %d",&n,&m))
	{
		for (int i=1;i<=n;i++)
		{
			f[i]=i;
			sum[i]=0;
		}
		ans=0;
		while (m--)
		{
			scanf ("%d %d %d",&x,&y,&l);
			fx=find(x);
			fy=find(y);
			if (fx!=fy)		//根节点不同，合并 
			{
				f[fy]=fx;
				sum[fy]=sum[x]-sum[y]+l;		//这里交换的时候还是要注意，虽然以前AC过，现在还是弄不太清 
			}
			else		//根节点相同，判断语句是否与前面出现的冲突 
			{
				if (l!=sum[y]-sum[x])
					ans++;
			}
		}
		printf ("%d\n",ans);
	}
	return 0;
}