P5787 线段树分治+可撤销并查集

• 题目链接：https://www.luogu.com.cn/problem/P5787
• 解题技巧：
  1.对于此类在某一段时间**[L,R]**内产生贡献的非强制在线问题，我们都可以试着采用线段树分治解决

//#define LOCAL
#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define mem(a, b) memset(a,b,sizeof(a))
#define sz(a) (int)a.size()
#define INF 0x3f3f3f3f
#define DNF 0x7f
#define DBG printf("this is a input\n")
#define fi first
#define se second
#define mk(a, b) make_pair(a,b)
#define pb push_back
#define LF putchar('\n')
#define SP putchar(' ')
#define p_queue priority_queue
#define CLOSE ios::sync_with_stdio(0); cin.tie(0)
#define sz(a) (int)a.size()
#define pii pair <int,int>
template<typename T>
void read(T &x) {x = 0;char ch = getchar();ll f = 1;while(!isdigit(ch)){if(ch == '-')f *= -1;ch = getchar();}while(isdigit(ch)){x = x * 10 + ch - 48; ch = getchar();}x *= f;}
template<typename T, typename... Args>
void read(T &first, Args& ... args) {read(first);read(args...);}
template<typename T>
void write(T arg) {T x = arg;if(x < 0) {putchar('-'); x =- x;}if(x > 9) {write(x / 10);}putchar(x % 10 + '0');}
template<typename T, typename ... Ts>
void write(T arg, Ts ... args) {write(arg);if(sizeof...(args) != 0) {putchar(' ');write(args ...);}}
using namespace std;

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    return a / gcd(a, b) * b;
}

const int N = 2e5 + 5;
int n , m , k;
vector <int> v[N << 2];
struct node
    {
    int u , v;
}edge[N];
struct Undo_Dsu
{
    stack <pii> st ;
    int fa[N] , siz[N] ;
    int d[N];
    void init()
    {
        while(!st.empty())  st.pop() ;
        for (int i = 1 ; i <= n  ; i ++)  fa[i] = i , siz[i] = 1 , d[i] = 0;
    }
    int dis (int x)
    {
        int temp = 0;
        while (fa[x] != x) {
            temp ^= d[x];
            x = fa[x];
        }
        return temp;
    }
    int findroot(int x)
    {
        return x == fa[x] ? x : findroot(fa[x]) ;
    }
    bool merge(int u , int v)
    {
        int fax = findroot(u) , fay = findroot(v);
        if(fax == fay)  return 0 ;
        if (siz[fax] > siz[fay])  swap(fax, fay) , swap(u , v) ;
        int d1 = dis (u), d2 = dis (v);
        fa[fax] = fay , siz[fay] += siz[fax] ;
        d[fax] = (d1 + d2 + 1) % 2;
        st.push({fax , fay}) ;
        return 1 ;
    }
    void undo() //撤销一次最新的合并
    {
        pii no = st.top();
        fa[no.fi] = no.fi ;
        siz[no.se] -= siz[no.fi] ;
        d[no.fi] = 0;
        st.pop() ;
    }
} dsu ;


void insert (int i , int L , int R, int l , int r , int val)
{
    if (L > r || R < l) return ;
    if (l <= L && R <= r) {
        v[i].pb(val);
        return ;
    }
    int mid = (L + R) >> 1;
    insert (i << 1 , L , mid , l , r , val);
    insert (i << 1 | 1 , mid + 1 , R , l , r, val);
}
void dfs (int i , int L, int R)
{
    int ok = 1;
    int t = sz(dsu.st);
    for (auto it : v[i])
    {
        int u = edge[it].u , v = edge[it].v;
        int fau = dsu.findroot (u) , fav = dsu.findroot (v);
        if (fau == fav)
        {
            int d1 = dsu.dis(u), d2 = dsu.dis(v);
            if ((d1 + d2) % 2 == 0) {
                ok = 0;
                break;
            }
        }
        else
            dsu.merge(u, v);
    }
    if (ok == 0)
    {
        for (int t = L ; t <= R ; t ++)
            cout << "No" << endl;
    }
    else
    {
        if (L == R)
        {
            cout << "Yes" << endl;
            return ;
        }
        int mid = (L + R) >> 1;
        dfs (i << 1 , L , mid);
        dfs (i << 1 | 1, mid + 1 , R);
    }
    while (sz(dsu.st) > t) dsu.undo();
    return ;
}
int main ()
{
    read (n , m, k);
    for (int i = 1 ; i <= m ; i ++)
    {
        int u , v, l , r;
        read (u , v ,l , r);
        edge[i].u = u, edge[i].v = v;
        insert (1 , 1 , k , l + 1 , r , i);
    }
    dsu.init();
    dfs (1, 1 , k);
}

本文地址：https://blog.csdn.net/Zenith_Habitant/article/details/109249873