AcWing 239.奇偶游戏 (带权并查集/种类并查集)

• 

• 题意:你和朋友玩游戏,有个一\(01\)序列,你每次给出一个区间,朋友会回答这个区间中的\(1\)的个数是奇数还是偶数,但是你亲爱的朋友可能在撒谎,问在哪个询问你能确定你的朋友在撒谎,输出回合数.

• 题解:假如区间\([l,r]\)所含的奇数个数为偶数的话,那么其前缀和\(s_{l-1}\)和\(s_r\)所含的\(1\)的个数一定同奇同偶,如果\([l,r]\)所含奇数个数为奇数,\(s_{l-1}\)和\(s_r\)奇偶性一定不同.

  所以我们对前缀和\(s\)进行维护,如果\([l,r]\)为偶数,那么我们可以将区间\(s_{l-1}\)和\(s_r\)合并,并且它们之间的权值应该为\(0\),若为奇数则权值为\(1\),传递关系可以用异或来操作,核心思想依然是带权并查集,但是这题还需要离散化.

  此处顺便附上种类并查集的做法

• 代码:

  #include <iostream>
  #include <iomanip>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define db double
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  int gcd(int a,int b){return b?gcd(b,a%b):a;}
  int lcm(int a,int b){return a/gcd(a,b)*b;}
  inline int read()
  {
      int X=0; bool flag=1; char ch=getchar();
      while(ch<'0'|ch>'9') {if(ch=='-') flag=0; ch=getchar();}
      while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
      if(flag) return X;
      return ~(X-1);
  }
  int n;
  int m;
  int a,b;
  string op;
  int p[N];
  int d[N];
  unordered_map<int,int> S;
  int get(int x){
  	if(S.count(x)==0) S[x]=++n;
  	return S[x];
  }
  int find(int x){
  	if(p[x]!=x){
  		int root=find(p[x]);
  		d[x]^=d[p[x]];
  		p[x]=root;
  	}
  	return p[x];
  }
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      rep(i,1,10010) p[i]=i;
      cin>>n>>m;
      n=0;
      int ans=m;
      rep(i,1,m){
      	cin>>a>>b>>op;
     		a=get(a-1),b=get(b);
      	int fa=find(a);
      	int fb=find(b);
      	int t=0;
      	if(op=="odd") t=1;
      	if(fa==fb){
      		if((d[a]^d[b])!=t){
      			ans=i-1;
      			break;
      		}
      	}
      	else{
      		p[fa]=fb;
      		d[fa]=d[a]^d[b]^t;
      	}
      }
      cout<<ans<<'\n';
      return 0;
  }
  **************************************************************
  #include <iostream>
  #include <iomanip>
  #include <cstdio>
  #include <cstring>
  #include <cmath>
  #include <algorithm>
  #include <stack>
  #include <queue>
  #include <vector>
  #include <map>
  #include <set>
  #include <unordered_set>
  #include <unordered_map>
  #define ll long long
  #define db double
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  int gcd(int a,int b){return b?gcd(b,a%b):a;}
  int lcm(int a,int b){return a/gcd(a,b)*b;}
  inline int read()
  {
      int X=0; bool flag=1; char ch=getchar();
      while(ch<'0'|ch>'9') {if(ch=='-') flag=0; ch=getchar();}
      while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();}
      if(flag) return X;
      return ~(X-1);
  }
  int n,m;
  int p[N];
  int a,b;
  string op;
  unordered_map<int,int> S;
  int get(int x){
  	if(S.count(x)==0) S[x]=++n;
  	return S[x];
  }
  int find(int x){
  	if(p[x]!=x) p[x]=find(p[x]);
  	return p[x];
  }
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      cin>>n>>m;
      int cnt=10010/2;
      n=0;
      rep(i,1,10010) p[i]=i;
      int ans=m;
      rep(i,1,m){
      	cin>>a>>b>>op;   //p[x]存偶数,p[x+n]存奇数
      	a=get(a-1),b=get(b);
      	if(op=="even"){
      		if(find(a+cnt)==find(b)){
      			ans=i-1;
      			break;
      		}
      		p[find(a)]=find(b);
      		p[find(a+cnt)]=find(b+cnt);
      	}
      	else{
      		if(find(a)==find(b)){
      			ans=i-1;
      			break;
      		}
      		p[find(a)]=find(b+cnt);
      		p[find(a+cnt)]=find(b);
      	}
      }
      cout<<ans<<'\n';
      return 0;
  }