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

题解：就是判断错误的条件数量，这种情况就是当两点的距离知道了（两点属于同一个集合就知道距离了），条件给出错误的答案。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int pre[500005];
int d[500005];

int find(int x)
{
	if(x == pre[x])
	{
		return x;
	}
	int fa = pre[x];
	pre[x] = find(pre[x]);
	d[x] += d[fa];
	return pre[x];
}

int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m) != EOF)
	{
		for(int i = 1;i <= n;i++)
		{
			pre[i] = i;
			d[i] = 0;
		}
		int res = 0;
		for(int i = 0;i < m;i++)
		{
			int A,B,X;
			scanf("%d%d%d",&A,&B,&X);
			int x = find(A);
			int y = find(B);
			if(x == y)
			{
				if(X != d[B] - d[A])
				{
					res++;
				}
			}
			else
			{
				pre[y] = x;
				d[y] = X - d[B] + d[A];  //两点根之间的距离，都是求右边到左边的距离
			}
		}

		printf("%d\n",res);
	}

	return 0;
}

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