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;}