[poj1703]Find them, Catch them(种类并查集)

题意：食物链的弱化版本

解题关键：种类并查集，注意向量的合成。

$rank$为1代表与父亲对立，$rank$为0代表与父亲同类。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath> 
using namespace std;
typedef long long ll;
#define M 100005
int fa[M],rank1[M];
int find1(int x){
    if(x==fa[x]) return x;
    int tmp=fa[x];
    fa[x]=find1(tmp);
    rank1[x]=(rank1[x]+rank1[tmp])%2;
    return fa[x];
}
void unite(int x,int y){
    int t=find1(x);
    int t1=find1(y);
    if(t!=t1){//合并的时候需要用到向量的合成和差 
        fa[t1]=t;
        rank1[t1]=(1-rank1[y]+rank1[x])%2;
    }
}
char ch[10];
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=0;i<=n;i++){
            fa[i]=i;
            rank1[i]=0;
        }
        for(int i=0;i<m;i++){
            int tmp,tmp1;
            scanf("%s%d%d",ch,&tmp,&tmp1);
            if(ch[0]=='D') unite(tmp,tmp1);
            else{
                int x=find1(tmp);
                int y=find1(tmp1);
                if(x==y){//可以判断出关系 
                    int r=(2-rank1[tmp]+rank1[tmp1])%2;
                    if(r==0) printf("In the same gang.
");
                    else printf("In different gangs.
");
                }
                else printf("Not sure yet.
");
            }
        }
    }
    return 0;
}

 法二：$fa$数组代表$i$属于$A$或$i$属于$B$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
using namespace std;
const int maxn=100005;
int fa[maxn*2];
int find1(int x){
    int r=x;
    while(r!=fa[r])  r=fa[r];
    int i=x,j;
    while(i!=r){
        j=fa[i];
        fa[i]=r;
        i=j;
    }
    return r;
}
 
void unite(int x,int y){
    x=find1(x),y=find1(y);
    if(x!=y) fa[x]=y;
}
 
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        int N,M,x,y;
        char opt[10];
        scanf("%d%d",&N,&M);
        for(int i=0;i<=2*N;i++) fa[i]=i;
        while(M--){
            scanf("%s%d%d",opt,&x,&y);
            if(opt[0]=='A'){
                if(find1(x)==find1(y)) printf("In the same gang.
");
                else if(find1(x)==find1(y+N)&&find1(x+N)==find1(y)) printf("In different gangs.
");
                else printf("Not sure yet.
");
            }
            else{
                unite(x,y+N);
                unite(x+N,y);
            }
        }
    }
    return 0;
}