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

Sample Output

2
Hint

Hint:
（PS： the 5th and 10th requests are incorrect）
这题可以用并查集做，属于带权值的并查集。t1,t2是根节点，其中root[t2]=root[a]+c-root[b];（可以用向量推）t2是集合二的根节点，a,b是输入的点
这里抄附上一张图
     root[a]
a--------ra
|             -
|x                   -
|         root[b]       -
b--------------------rb
原本以为rb经过集合合并后根节点是a,但其实是ra，所以可以递归并查集算出每个节点离根节点的距离

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
int pre[50005],root[50005];
int find(int x)
{
	int temp;
	if(x==pre[x])return x;
	temp=pre[x];
	pre[x]=find(pre[x]);
	root[x]=root[x]+root[temp];
	return pre[x];
}


int main()
{
	int n,m,i,j,a,b,c,t1,t2,ans;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<=n;i++){
			pre[i]=i;root[i]=0;
		}
		ans=0;
		for(i=1;i<=m;i++){
			scanf("%d%d%d",&a,&b,&c);
			t1=find(a);t2=find(b);
			if(t1==t2){
				if(root[b]-root[a]!=c){
					ans++;continue;
				}
			}
			else{
				pre[t2]=t1;
				root[t2]=root[a]+c-root[b];
			}
		}
		printf("%d
",ans);
	}
	return 0;
}