poj1182 拆点并查集

poj1182

题意：中文题

思路：去年做的带权并查集，拆点的姿势还是要学习一个的

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a) memset(a,0,sizeof(a))
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e5+100;
const ll mod=1e9+7;

int n,k,ans,pre[N<<1];
void init(int n){
    for(int i=0; i<=3*n; ++i) pre[i]=i;
}
int finds(int x){
    return pre[x]=pre[x]==x?x:finds(pre[x]);
}
void unions(int x, int y){
    int fx=finds(x), fy=finds(y);
    pre[fy]=fx;
}
int main(){
    //ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    cin>>n>>k;
    init(n);
    for(int i=1; i<=k; ++i){
        int d,x,y;
        scanf("%d%d%d",&d,&x,&y);
        if(x>n || x<1 || y>n || y<1){
            ans++;
            continue;
        }
        if(d==1){
            if(finds(x)==finds(y+n) || finds(x)==finds(y+2*n)){
                ans++;
            }
            else{
                unions(x,y);
                unions(x+n,y+n);
                unions(x+2*n,y+2*n);
            }
        }
        else{
            if(finds(x)==finds(y) || finds(x)==finds(y+2*n)){
                ans++;
            }
            else{
                unions(x,y+n);
                unions(x+n,y+2*n);
                unions(x+2*n,y);
            }
        }
    }
    cout<<ans<<"
";
    return 0;
}