傳送門：http://www.lydsy.com/JudgeOnline/problem.php?id=4025

【題解】

考慮對時間分治，用可撤回的啟發式合並並查集來維護連通性。

二分圖的條件是沒有奇環，用並查集判即可。

對於時間區間[l,r]，如果邊在這期間都存在，那麽就加入並查集，對於剩下的邊分類，並且分治下去做。

對於每條邊，在log個區間表示出來了，需要進行判斷，啟發式合並的getf是log的，所以復雜度為O（nlog^2n）

# include <stdio.h>
# include <string.h>
# include <iostream>
# include 
 
<algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, N = 1e6 + 10;
const int mod = 1e9+7;

# define RG register
# define ST static

int n, m, T;

struct edge {
    int u, v, st, ed;
    edge() {}
    edge(
 
int u, int v, int st, int ed) : u(u), v(v), st(st), ed(ed) {}
}e[M]; 

int st[N];
int stn = 0; 

struct us {
    int n, fa[M], rk[M], dis[M];
    inline void set(int _n) {
         n = _n;
         for (int i=1; i<=n; ++i) fa[i] = i, rk[i] = 1, dis[i] = 0;
    }
    inline int getf(int x) {
        
 
return fa[x] == x ? x : getf(fa[x]); 
    }
    inline int getdis(int x) {
        int d = 0;
        while(fa[x] != x) d += dis[x], x = fa[x]; 
        return d;
    }
    inline void un(int fu, int fv, int d) {
        if(rk[fu] > rk[fv]) swap(fu, fv);
        if(rk[fu] == rk[fv]) rk[fv] ++, st[++stn] = -fv; 
        fa[fu] = fv;
        dis[fu] = d; 
        ++stn;
        st[stn] = fu;
    }
    inline void re() {
        int x = st[stn]; --stn;
        if(x < 0) --rk[-x];
        else dis[x] = 0, fa[x] = x;
    }
}E;
        
bool ans[M]; 

inline void solve(int tl, int tr, int er) {
//    printf("l = %d, r = %d\n", tl, tr); 
    int cur_stn = stn;
    for (int i=1; i<=er; ++i) {
        if(!(e[i].st <= tl && tr <= e[i].ed)) continue; 
//        printf("doing: (%d, %d, %d, %d)\n", e[i].u, e[i].v, e[i].st, e[i].ed); 
        int fu = E.getf(e[i].u), fv = E.getf(e[i].v);
        int du = E.getdis(e[i].u), dv = E.getdis(e[i].v);
        if(fu != fv) E.un(fu, fv, du+dv+1);
        else {
//            printf("%d %d fa = %d %d %d\n", e[i].u, e[i].v, E.fa[1], du, dv); 
            if((du+dv+1)&1) {
//                puts("Proves odd"); 
                while(stn != cur_stn) E.re();
                return ;
            }
        }
        swap(e[i], e[er]); --er; --i; 
    }
    if(tl == tr) {
        ans[tl] = 1;
        while(stn != cur_stn) E.re(); 
        return ;
    }
    int mid = tl+tr >> 1;
    int ER = er;
    for (int i=1; i<=er; ++i) 
        if(e[i].st > mid) {
            swap(e[i], e[er]); 
            --er; --i;
        }
    solve(tl, mid, er);
    er = ER;
    for (int i=1; i<=er; ++i) {
        if(e[i].ed <= mid) {
            swap(e[i], e[er]);
            --er; --i;
        }
    }
    solve(mid+1, tr, er); 
    while(stn != cur_stn) E.re(); 
}

int main() {
    cin >> n >> m >> T;
    E.set(n); 
    int en = 0;
    for (int i=1, u, v, sta, end; i<=m; ++i) {
        scanf("%d%d%d%d", &u, &v, &sta, &end);
        ++sta; if(sta>end) continue;
        e[++en] = edge(u, v, sta, end);
    }
    m = en;
    solve(1, T, m); 
    for (int i=1; i<=T; ++i) puts(ans[i] ? "Yes" : "No"); 
    return 0;
}

bzoj4025 二分圖