HDU3047 Zjnu Stadium【种类并查集】

Zjnu Stadium

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5612    Accepted Submission(s): 2161

 

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

Hint: （PS： the 5th and 10th requests are incorrect）

Source

2009 Multi-University Training Contest 14 - Host by ZJNU

问题链接：HDU3047 Zjnu Stadium

问题描述：在一个体育馆内，有300列环型座位（顺时针从1开始编号），假定每一列有无数个座位，现在想为N个人预定座位，A B X表示B的座位在A的的座位顺时针第X个座位，对于若干组A B X问与前面的预定有冲突的个数

解题思路：种类并查集，pre存储i的父节点 pos存储i到父节点的距离，距离要模300，0表示位置300。具体修改看代码

AC的C++代码：

#include

using namespace std;

const int N=50010;
const int MOD=300;

int pre[N],pos[N];//pre存储i的父节点 pos存储i到父节点的距离，0表示位置300
void init(int n)
{
	for(int i=0;i<=n;i++){
		pre[i]=i;//自己为自己的父节点 
		pos[i]=0;//自己到自己的距离为0 
	}
}

int find(int x)
{
	if(x!=pre[x]){
		int r=pre[x];
		pre[x]=find(r);
		pos[x]=(pos[x]+pos[r])%MOD;
	}
	return pre[x];
}

bool join(int x,int y,int d)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy){//fx和fy是两颗树，合并 
		pre[fy]=fx;
		pos[fy]=(MOD+pos[x]-pos[y]+d)%MOD;
	}
	else{
		if((pos[x]+d)%MOD!=pos[y])//出现冲突 
		  return true;
	}
	return false;
}

int main()
{
	int n,m,a,b,x;
	while(~scanf("%d%d",&n,&m)){
		init(n);
		int ans=0;
		while(m--){
			scanf("%d%d%d",&a,&b,&x);
			if(join(a,b,x))
			  ans++;
		}
		printf("%d\n",ans);
	}
	return 0;
}