Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 198    Accepted Submission(s): 86

 

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

10 101 2 1503 4 2001 5 2702 6 2006 5 804 7 1508 9 1004 8 501 7 1009 2 100

Sample Output

2
Hint
Hint:（PS： the 5th and 10th requests are incorrect）
 
此题为带权并查集，注意对距离取模。
 
#include<cstdio>using namespace std;const int MAXN = 50000 + 10;int p[MAXN], total[MAXN];  // total表示距离其最顶层的祖先的距离 int n, m;void init(){for(int i = 1; i <= n; i++){p[i] = i;total[i] = 0;}}int find(int x){if(x != p[x]){int temp = p[x];p[x] = find(p[x]);  //路径压缩 total[x] += total[temp];total[x] %= 300; }return p[x];}bool join(int a, int b, int c){int x = find(a);int y = find(b);if(x == y){if((total[a] + c) % 300 != total[b]) return false;return true;}p[y] = x;total[y] = (total[a] + c - total[b] + 300) % 300;  //环形座位,+300保证为正数 return true;}int main(){int a, b, c, cnt;while(~scanf("%d%d", &n, &m)){cnt = 0;init();for(int i = 1; i <= m; i++){scanf("%d%d%d", &a, &b, &c);if(!join(a, b, c)) cnt++;}printf("%d\n", cnt);}return 1;}