mirror of
https://github.com/Kiritow/OJ-Problems-Source.git
synced 2024-03-22 13:11:29 +08:00
62 lines
1.1 KiB
C++
62 lines
1.1 KiB
C++
|
|
|||
|
|
#include<iostream>
|
|||
|
|
#include<hash_map>
|
|||
|
|
//#include"iphxer_h.h"
|
|||
|
|
#include<string>
|
|||
|
|
#include<cstring>
|
|||
|
|
#include<cstdio>
|
|||
|
|
#include<map>
|
|||
|
|
|
|||
|
|
using namespace std;
|
|||
|
|
#define maxsize 50100
|
|||
|
|
int rank1[maxsize];//相对于根元素的权值0：同类，1：x吃father[x],2表示被father[x] 吃
|
|||
|
|
int father[maxsize];
|
|||
|
|
void initU(int n){
|
|||
|
|
for(int i=0;i<n;i++){
|
|||
|
|
father[i]=i;
|
|||
|
|
rank1[i]=0;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
int Find(int x){
|
|||
|
|
if(father[x]==x)return x;
|
|||
|
|
int t=father[x];//保存更新前的father位置，用来更新子rank
|
|||
|
|
father[x]=Find(father[x]);
|
|||
|
|
rank1[x]=(rank1[t]+rank1[x])%3;//计算相对于根节点的rank值
|
|||
|
|
return father[x];
|
|||
|
|
}
|
|||
|
|
//把y合并到x中，关系为d
|
|||
|
|
void Union(int x,int y,int d)
|
|||
|
|
{
|
|||
|
|
int xf=Find(x);
|
|||
|
|
int yf=Find(y);
|
|||
|
|
father[xf]=yf;//合并到y上
|
|||
|
|
rank1[xf]=(rank1[y]-rank1[x]+3+d)%3;
|
|||
|
|
}
|
|||
|
|
|
|||
|
|
int main()
|
|||
|
|
{
|
|||
|
|
|
|||
|
|
int i,n,m;
|
|||
|
|
int d,x,y,xf,yf;
|
|||
|
|
int sum=0;
|
|||
|
|
scanf("%d%d",&n,&m);
|
|||
|
|
initU(n);
|
|||
|
|
for(int i=0;i<m;i++){
|
|||
|
|
scanf("%d%d%d",&d,&x,&y);
|
|||
|
|
if(x>n||y>n||(d==2&&x==y))
|
|||
|
|
++sum;
|
|||
|
|
else{
|
|||
|
|
xf=Find(x);
|
|||
|
|
yf=Find(y);
|
|||
|
|
if(xf==yf){
|
|||
|
|
if((rank1[x]-rank1[y]+3)%3!=d-1)
|
|||
|
|
sum++;
|
|||
|
|
}else{
|
|||
|
|
Union(x,y,d-1);
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
cout<<sum<<endl;
|
|||
|
|
}
|
|||
|
|
|