Problem W

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.<br>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).<br>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.<br>

 

Input

There are many test cases:<br>For every case: <br>The first line has two integer N(1<=N<=50,000), M(0<=M<=100,000),separated by a space.<br>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.<br><br>

 

Output

For every case: <br>Output R, represents the number of incorrect request.<br>

 

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<div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div>Hint:（PS： the 5th and 10th requests are incorrect）</div>

简单题意：

  有编号为1-N的人在体育馆里，输入为 A B X ,代表二者之间的距离为X。然后给出一组数据，求出有多少错误的数据，即有冲突的数据，当前的关系与之前的关系有冲突。

解题思路形成过程：

  运用并查集求出带权的树，这是核心的思想。并查集的代码模板可以完全胜任这个程序，所以问题也就自然而然的解决。关键是并查集的运用。

感想：

  图论这个专题，其实新颖的东西并不多，只要掌握了核心，一招鲜吃遍天下。

AC代码：

#include<cstdio>
#include<cmath>
using namespace std;
#define N 50005
int f[N], rank[N], n, m;

void init(){
    for(int i=0; i<=n; ++i)
        f[i]=i, rank[i]=0;
}
int find(int x){
    if(x==f[x]) return f[x];
    int t=f[x];
    f[x] = find(f[x]);
    rank[x] += rank[t];
    return f[x];
}
bool Union(int x,int y, int m){
    int a=find(x), b=find(y);
    if(a==b){
        if(rank[x]+m!=rank[y])
            return false;
        return true;
    }
    f[b] = a;
    rank[b] = rank[x]+m-rank[y];
    return true;
}

int main(){
    int a,b,x;
    while(~scanf("%d%d",&n,&m)){
        init();
        int cnt=0;
        for(int i=0; i<m; ++i){
            scanf("%d%d%d",&a,&b,&x);
            if(!Union(a, b, x)){
                ++cnt;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}