poj 1182 食物链

题目链接:http://poj.org/problem?id=1182

题目大意:

中文,题意不解释...

题目思路:

并查集---食物链问题.

dis[i]=0 : i与根同类.

dis[i]=1 : i吃根.

dis[i]=2 : i被根吃.

ps:比较喜欢用非递归的方法写并查集...个人爱好.

代码:

#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <ctype.h>
#include <math.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll long long
#define ls rt<<1
#define rs ls|1
#define lson l,mid,ls
#define rson mid+1,r,rs
#define middle (l+r)>>1
#define eps (1e-8)
#define type int
#define clr_all(x,c) memset(x,c,sizeof(x))
#define clr(x,c,n) memset(x,c,sizeof(x[0])*(n+1))
#define MOD 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define _max(x,y) (((x)>(y))? (x):(y))
#define _min(x,y) (((x)<(y))? (x):(y))
#define _abs(x) ((x)<0? (-(x)):(x))
#define getmin(x,y) (x= ((x)<0 || (y)<(x))? (y):(x))
#define getmax(x,y) (x= ((y)>(x))? (y):(x))
template <class T> void _swap(T &x,T &y){T t=x;x=y;y=t;}
int TS,cas=1;
const int M=50000+5;
int n,k;
int fa[M],dis[M];

int find(int x){
	int rt,i,j,sum=0;
	for(rt=x;rt!=fa[rt];rt=fa[rt]) sum+=dis[rt];
	for(i=x;i!=rt;i=j){
		sum-=dis[i];
		dis[i]=(dis[i]+sum)%3;
		j=fa[i];
		fa[i]=rt;
	}
	return rt;
}

void run(){
	scanf("%d%d",&n,&k);
    int i,j;
	for(i=1;i<=n;i++) fa[i]=i,dis[i]=0;
	int cnt=0,o,x,y,xf,yf;
	while(k--){
		scanf("%d%d%d",&o,&x,&y);
		if(x>n || y>n || (o==2 && x==y)) cnt++;
		else{
			xf=find(x),yf=find(y);
			if(o==1) o=0;
			if(xf==yf){
				if((dis[x]+o)%3!=dis[y]) cnt++;
			}else{
				fa[yf]=xf;
				dis[yf]=(o-dis[y]+3+dis[x])%3;
			}
		}
	}
	printf("%d\n",cnt);
}

void preSof(){
}

int main(){
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    preSof();
    run();
    //while(~scanf("%d%d",&n,&k)) run();
    //for(scanf("%d",&TS);cas<=TS;cas++) run();
    return 0;
}