HDU3047 Zjnu Stadium
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:
（PS： the 5th and 10th requests are incorrect）
就是有一个环，给出多组关系满足a=b+x，判断不满足的有多少组
然后带权并查集维护一下，主义在合并两个并查集的时候差量的计算
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 50010 4 int fa[N],dis[N]; 5 int n,m; 6 int Find(int x){ 7 if(fa[x]==x)return x; 8 int f=Find(fa[x]); 9 dis[x]+=dis[fa[x]]; 10 return fa[x]=f; 11 } 12 void init(){for(int i=1;i<=n;i++)fa[i]=i,dis[i]=0;} 13 int main(){ 14 while(scanf("%d%d",&n,&m)!=EOF){ 15 init(); 16 int res=0; 17 while(m--){ 18 int a,b,x; 19 scanf("%d%d%d",&a,&b,&x); 20 int f_a=Find(a); 21 int f_b=Find(b); 22 if(f_a==f_b){ 23 if(dis[a]+x!=dis[b])res++; 24 continue; 25 } 26 dis[f_b]=dis[a]-dis[b]+x; 27 fa[f_b]=f_a; 28 } 29 printf("%d ",res); 30 } 31 return 0; 32 }