[CEOI1999] Parity Game（并查集）

方法1：带权路径维护
本题核心：[a,b]之间有奇数个1转换为s[a-1]^s[b] = 1，从而转向并查集

#include<bits/stdc++.h>
using namespace std;
#define x first
#define y second
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef vector<string> VS;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
vector<int> vx;
inline void divide() {sort(vx.begin(),vx.end());vx.erase(unique(vx.begin(),vx.end()),vx.end());}
inline int mp(int x) {return upper_bound(vx.begin(),vx.end(),x)-vx.begin();}
inline int log_2(int x) {return 31-__builtin_clz(x);}
inline int popcount(int x) {return __builtin_popcount(x);}
inline int lowbit(int x) {return x&-x;}
const int N = 5010;
//d[i]表示i到根节点的异或值是1还是0
int f[N<<1],d[N<<1],sz[N<<1];
int find(int u)
{
    if(f[u] == u) return u;
    int t = find(f[u]);
    d[u] ^= d[f[u]];
    return f[u] = t;
}
void merge(int u,int v,int st)
{
    int f1 = find(u), f2 = find(v);
    //要求d[u]^d[f1]^d[v] = st,反推d[f1]
    f[f1] = f2, d[f1]^=(d[u]^d[v]^st);
}
void solve()
{
    int n;
    cin>>n;
    int m;
    cin>>m;
    for(int i=1;i<=m*2;++i) f[i] = i;
    vector<array<int,3>> a(m);
    for(int i=0;i<m;++i)
    {
        string op;
        cin>>a[i][0]>>a[i][1]>>op;
        //[a,b]中有奇数个1即s[a-1]和s[b]的奇偶性不同
        vx.push_back(a[i][0]-1);
        vx.push_back(a[i][1]);
        if(op == "even") a[i][2] = 0;
        else a[i][2] = 1;
    }
    //注意离散化需要erase掉重复的内容
    divide();
    for(int i=0;i<m;++i)
    { 
        int u = mp(a[i][0]-1), v = mp(a[i][1]), st = a[i][2];
        //cout<<u<<' '<<v<<' '<<st<<'\n';
        if(find(u) == find(v))
        {
            if(d[u]^d[v] != st) 
            {
                cout<<i<<'\n';return ;
            }
        }
        else
        {
            merge(u,v,st);
        }
    }
    cout<<m<<'\n';
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T = 1;
	//cin>>T;
	while(T--)
	{
		solve();
	}
}

方法2扩展域

#include<bits/stdc++.h>
using namespace std;
#define x first
#define y second
typedef pair<int,int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
typedef vector<string> VS;
typedef vector<int> VI;
typedef vector<vector<int>> VVI;
vector<int> vx;
inline void divide() {sort(vx.begin(),vx.end());vx.erase(unique(vx.begin(),vx.end()),vx.end());}
inline int mp(int x) {return upper_bound(vx.begin(),vx.end(),x)-vx.begin();}
inline int log_2(int x) {return 31-__builtin_clz(x);}
inline int popcount(int x) {return __builtin_popcount(x);}
inline int lowbit(int x) {return x&-x;}
const int N = 5010;
//d[i]表示i到根节点的异或值是1还是0
int f[N<<2];
int find(int u)
{
    if(f[u] == u) return u;
    return f[u] = find(f[u]);
}
void merge(int u,int v)
{
    int f1 = find(u), f2 = find(v);
    f[f1] = f2;
}
void solve()
{
    int n;
    cin>>n;
    int m;
    cin>>m;
    //拆点
    for(int i=1;i<=m*4;++i) f[i] = i;
    vector<array<int,3>> a(m);
    for(int i=0;i<m;++i)
    {
        string op;
        cin>>a[i][0]>>a[i][1]>>op;
        //[a,b]中有奇数个1即s[a-1]和s[b]的奇偶性不同
        vx.push_back(a[i][0]-1);
        vx.push_back(a[i][1]);
        if(op == "even") a[i][2] = 0;
        else a[i][2] = 1;
    }
    //注意离散化需要erase掉重复的内容
    divide();
    //用i*2表示i的偶数点,用i*2-1表示i的奇数点,表示sum[i]是奇数或者偶数
    for(int i=0;i<m;++i)
    { 
        int u = mp(a[i][0]-1), v = mp(a[i][1]), st = a[i][2];
        int ue = u*2, uo = u*2-1, ve = v*2, vo = v*2-1;
        //若st == 1则uo应该和ve合并，ue和vo合并
        if(st)
        {
            if(find(uo) == find(vo)||find(ue) == find(ve))
            {
                cout<<i<<'\n';return ;
            }
            merge(uo,ve);merge(ue,vo);
        }
        else
        {
            if(find(ue) == find(vo)||find(uo) == find(ve))
            {
                cout<<i<<'\n';return ;
            }
            merge(uo,vo);merge(ue,ve);
        }
    }
    cout<<m<<'\n';
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T = 1;
	//cin>>T;
	while(T--)
	{
		solve();
	}
}