杭电 oj3047~带权并查集

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2947    Accepted Submission(s): 1132

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）

 

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

 

#include<iostream>
#include<cstdio>
using namespace std;

#define maxsize 50050

int pre[maxsize];
int dist[maxsize];//记录每个点到根的距离，初始值置为0

void init(int n)
{
    for(int i=1;i<=n;i++){pre[i]=i;dist[i]=0;}
}


int find(int x)
{
    if(x==pre[x]) return x;
    int tmp=find(pre[x]);//进行回溯更新每个节点的dist值
    dist[x]+=dist[pre[x]];
    pre[x]=tmp;
    return tmp;//返回根
}


int Union(int u,int v,int w)
{
    int r1=find(u),r2=find(v);
    if(r1==r2&&dist[v]!=dist[u]+w) return 0;
    else{
        pre[r2]=r1;
        dist[r2]=dist[u]+w-dist[v]; //利用向量来推的数学公式
        return 1;
    }
}


int main()
{
    int u,v,w;int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        init(n);
        int ans=0;
        while(m--)
        {
            scanf("%d %d %d",&u,&v,&w);
            if(!Union(u,v,w)) {ans++;}
        }
        printf("%d
",ans);
    }
}