HDU - 3047 Zjnu Stadium

Posted 2020-10-02 西北会法语

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）

和食物链一个思路，用树表示的并查集

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 
 6 using namespace  std;
 7 
 8 struct node
 9 {
10     int par;
11     int re;
12 }M[50005];
13 
14 
15 int find(int x)
16 {
17     if(M[x].par==x)
18         return x;
19     int t=M[x].par;
20     M[x].par=find(t);
21     M[x].re=M[x].re+M[t].re;
22     return M[x].par;
23 }
24 
25 int main()
26 {
27     int N,K,sum=0,a,b,x,f1,f2;
28     while(~scanf("%d%d",&N,&K))
29     {
30         sum=0;
31         for(int i=1;i<=N;i++)
32         {
33             M[i].par=i;
34             M[i].re=0;
35         }
36         for(int i=1;i<=K;i++)
37         {
38             scanf("%d%d%d",&a,&b,&x);
39             f1=find(a);
40             f2=find(b);
41             if(f1!=f2)
42             {
43                 M[f2].par=f1;
44                 M[f2].re=M[a].re+x-M[b].re;//????
45                 //cout<<"~~~"<<f1<<" "<<f2<<" "<<M[f2].re<<endl;
46             }
47             else
48             {
49                 if((M[b].re-M[a].re)!=x)
50                 {
51                     sum++;
52                     continue;
53                 }
54             }
55         }
56         printf("%d\n",sum);
57     }
58     
59     
60     return 0;
61 }