Zjnu Stadium - 带权并查集

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. 

InputThere 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. 

OutputFor 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）

分析：这道题和How many answers are we wrong几乎一样，点击打开链接

唯一不同的是左端点不用减1了

例如，sum[i]表示i到根节点的距离，1为根节点，有4个点

序号：    1    2    3    4

位置：    3    10    4    6

x：         0    7      1   3

sum[4]-sum[3]=2表示4就是表示第4个到第2个的距离，不用减1

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,pre[50004],sum[50004];

int f(int x){
    if(pre[x]==x)return x;
    int tmp=pre[x];
    pre[x]=f(pre[x]);
    sum[x]=sum[x]+sum[tmp];
    return pre[x];
}

int main(){
    int a,b,x;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=0;i<=n;i++){
            pre[i]=i;
            sum[i]=0;
        }
        int ans=0;
        for(int q=1;q<=m;q++){
            scanf("%d%d%d",&a,&b,&x);
            int root1=f(a),root2=f(b);
            if(root1!=root2){//没有被连到一起
                pre[root2]=root1;
                sum[root2]=x+sum[a]-sum[b];
            }
            else{
                if(sum[b]-sum[a]!=x)ans++;
            }
        }
        printf("%d\n",ans);
    }
}