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): 989    Accepted Submission(s): 388

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）

    题意：

    给出N（1到50000）个人和M（0到1000000）条关系。每条关系有A,B,W，代表A位置与B位置有W个距离，A在前，B在后，全部位置最多有300个。给出M种关系中，判断错误的位置关系有几条。输出错误的信息数。

    

    思路：

    带权并查集。权值记录A到B的距离，如果判断A和B不在同一个根节点，则合并；如果在同一个根节点则查询判断是否为错误信息，是则ans增加1。

 

    AC：

#include<stdio.h>
#define max 50000+5
int root[max],re[max];
int find(int a)
{
	if(root[a]==a) return a;
	int r=find(root[a]);
	re[a]=(re[a]+re[root[a]])%300;
	root[a]=r;
	return r;
}

int main()
{
	int n,m,ans;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
	ans=0;
	for(int i=1;i<=n;i++)
	{
		root[i]=i;
		re[i]=0;
	}
	while(m--)
	{
		int a,b,w;
		int fa,fb;
		scanf("%d%d%d",&a,&b,&w);
		fa=find(a);
		fb=find(b);
		if(fa!=fb)
		{
			root[fa]=fb;
			re[fa]=(re[b]+w-re[a]+300)%300;
		}
		else
		{
			if((re[a]-re[b]+300)%300!=w) ans++;
		}
	}
	printf("%d\n",ans);
    }
	return 0;
}

 

   总结：

   WA了一次，因为输入要用EOF输入（There are many test cases）。