hdu 3047 Zjnu Stadium

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2109 Accepted Submission(s): 808


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）

這是大神的解析，原來是向量的原理：

 

（1）弄清題意，找出出現沖突的位置，判斷沖突很簡單就是當兩個人在同一行坐，同時他們到根節點的距離差值正好是他們之間的差值，此時就出現了沖突了。

（2）關鍵有兩個地方，這也是並查集題目的難點，就是壓縮集合，和求節點到根的距離。這裡壓縮集合就很簡單了，一個通用的遞歸。求到跟的距離dist[a] += dist[tem]; dist[rb]=dist[a]+x-dist[b];注意這兩行代碼，這是核心代碼，首先第一行是求出節點a到根的距離。第二行代碼使用的是數學中向量計算的原理如圖

圖

 

#include
#include
#define M 50000+10
int x[M],rank[M];//rank表示的是到根節點的距離 
int find(int k)
{
	int temp=x[k];
	if(x[k]==k) return k;
	x[k]=find(x[k]);
	rank[k]+=rank[temp];
	return x[k];
}
int merge(int a,int b,int m)
{
	int fa=find(a);
	int fb=find(b);
	if(fa==fb){
		if(rank[b]+m!=rank[a])//向量計算 
			return 0;
		else return 1;
	}
	x[fa]=fb;
	rank[fa]=rank[b]+m-rank[a];//用向量的運算法則計算 
	return 1;
}
int main()
{
	int n,m,i,a,b,d;
	while(~scanf("%d%d",&n,&m)){
		for(i=0;i