Zjnu Stadium HDU - 3047(带权并查集)-白红宇

Zjnu Stadium HDU - 3047(带权并查集)

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发布时间：2019-05-25

本文共 2413 字，大约阅读时间需要 8 分钟。

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

Hint

Hint:

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

一个体育馆内有300个座位，编号为B的人要在编号为A的人的顺时针方向x个位置。问给出的座位有多少个是错误的。

今晚做的这几个带权并查集的题目都是这样的，这个题目也不例外，但是这里注意是300个座位围成一个圈。所以需要取模300。其实不取模直接算也是可以的。但是取模之后要注意，在做差的时候有可能是负数，所以需要加上300。

代码如下：

#include
   
    #define ll long long#define mod 300using namespace std;const int maxx=5e4+100;int f[maxx],dis[maxx];int n,m;inline int getf(int u){
   	int x;	if(u!=f[u])	{
   		x=getf(f[u]);		dis[u]=(dis[u]+dis[f[u]])%mod;		f[u]=x;	}	return f[u];}inline int merge(int u,int v,int val){
   	int t1=getf(u);	int t2=getf(v);	if(t1==t2)	{
   		swap(u,v);		if((dis[u]+val)%mod!=dis[v]) return 1;		else return 0;	}	else	{
   		f[t1]=t2;		dis[t1]=(dis[v]-dis[u]+val+mod)%mod;//这一块要自己好好理解，容易写错了。。		return 0;	}}int main(){
   	int x,y,v;	while(~scanf("%d%d",&n,&m))	{
   		int ans=0;		for(int i=0;i<=n;i++) f[i]=i,dis[i]=0;		while(m--)		{
   			scanf("%d%d%d",&x,&y,&v);//swap(x,y);			if(merge(x,y,v)) ans++;		}		printf("%d\n",ans);	}	return 0;}

努力加油a啊，(^o)/~

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