HDU3047-Zjnu Stadium-带权并查集

Posted 2020-11-24 ofshk

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）

 1 #include<stdio.h>
 2 #include<map>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<algorithm>
 6 #include<stack>
 7 #include<cmath>
 8 #include<vector>
 9 using namespace std;
10 typedef long long ll;
11 const int N=50020;
12 
13 int f[N],d[N],n;
14 
15 void init()
16 {
17     for(int i=1;i<=n;i++)
18     {
19         f[i]=i;
20         d[i]=0;
21     }
22 }
23 int getf(int x)
24 {
25     if(f[x]==x)
26         return x;
27     else
28     {
29         int t=f[x];//找它的父节点
30         f[x]=getf(f[x]);
31         d[x]=d[x]+d[t];//d[x]（为x到根节点的距离）更新为它到它父节点的距离+父节点到根节点的距离
32         return f[x];
33     }
34 }
35 
36 void merge(int x,int y)
37 {
38     int t1=getf(x);
39     int t2=getf(y);
40     if(t1!=t2)
41         f[t2]=t1;
42     return;
43 }
44 
45 int main()
46 {
47     int m,x,y,z;
48     while(~scanf("%d %d",&n,&m))
49     {
50         init();
51         int ans=0;
52         for(int i=0;i<m;i++)
53         {
54             scanf("%d %d %d",&x,&y,&z);
55             int fx=getf(x),fy=getf(y);
56             if(fx==fy)
57             {
58                 int w=d[y]-d[x];
59                 if(w!=z)
60                     ans++;
61             }
62             else
63             {
64                 merge(x,y); //f[fy]=x;
65                 d[fy]=d[x]+z-d[y];
66             }
67         }
68         cout<<ans<<endl;
69     }
70     return 0;
71 }