Zjnu Stadium HDU

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

10 101 2 1503 4 2001 5 2702 6 2006 5 804 7 1508 9 1004 8 501 7 1009 2 100

Sample Output

2          

Hint

Hint:（PS： the 5th and 10th requests are incorrect）

题意：n个观众标号1~n，坐在环形的观众席上，一个位置可以坐好几个人。 给出 a b x 即b在a的顺时针x距离处。  现在给出你m条信息，让你判断错误的有几条（和前面冲突的就算错）

分析：  就我目前的水平看来，并查集实际上是一种维护森林关系的这么一种结构，每一个森林中的每一棵树都可以通过“母树（根）” 找到对应的关系，合并时只要将母树合并即可，原本两个森林里的任意两棵树都可以通过新的母树找到对应的关系。

就本题而言，这里的关系无非是两棵树之间的距离，那么我们如何确定两颗树的距离呢？ 就像我刚才说的，通过母树。 假设A到母树的距离为a，B到母树的距离为b（都是顺时针的距离）。 那么A到B的顺时针距离就是a-b。  

需要注意的是路径压缩的时候，容易错，想法是直接算出到母树的距离，但写法上要注意！具体看代码

代码：

#include<iostream>#include<cstdio>using namespace std;const int maxn = 50010;int par[maxn],val[maxn];int search(int a){    if(a == par[a]) return a;    int temp = par[a];    par[a] = search(par[a]);            //wa在这里    val[a] += val[temp];    val[a] %= 300;    return par[a] ;}void unite(int a,int b,int x){    int ra = search(a);    int rb = search(b);    par[ra] = rb;    val[ra] += (x+val[b]-val[a]+300);     //为了防止负数    val[ra] %= 300;}bool same(int a,int b){    return search(a) == search(b);}int main(){    int m,n;    while(scanf("%d%d",&n,&m)!=EOF){        int cnt = 0;        for(int i=0;i<=n;i++){            par[i] = i;            val[i] = 0;        }        for(int i=0;i<m;i++){            int a,b,x;            scanf("%d%d%d",&a,&b,&x);            if(same(a,b)){                if((val[a]-val[b]+300)%300 != x){                    cnt++;                }            }            else{                unite(a,b,x);            }        }        printf("%d\n",cnt);    }}