首先对时间分治，每个节点表示一个时间的区间$[l, r]$
然后对于每个节点记录一个可以撤销的并查集，维护图的信息即可
(这里的并查集不用路径压缩，只要按秩合并，这样子可以保证单次操作的时间复杂度是$O(logn)$的)
我去啊。。。把边分类这一段代码调的我QAQ了。。。代码能力太弱QAQQQ
1 /************************************************************** 2 Problem: 4025 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:4344 ms 7 Memory:6668 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 const int N = 1e5 + 5; 15 const int M = 2e5 + 5; 16 17 inline int read(); 18 19 struct edge { 20 int x, y, st, ed; 21 22 inline void get() { 23 x = read(), y = read(); 24 st = read() + 1, ed = read(); 25 } 26 } e[M]; 27 28 int n, m, time; 29 30 namespace set { 31 int fa[N], d[N], a[N]; 32 int s[N << 2], top = 0; 33 34 int find(int x) { 35 while (fa[x] != x) x = fa[x]; 36 return x; 37 } 38 int dep(int x) { 39 static int res; 40 for (res = 0; fa[x] != x; x = fa[x]) 41 res ^= a[x]; 42 return res; 43 } 44 45 void set_union(int x, int y, int _d) { 46 if (d[x] > d[y]) swap(x, y); 47 if (d[x] == d[y]) ++d[y], s[++top] = -y; 48 fa[x] = y, a[x] = _d, s[++top] = x; 49 } 50 51 void set_resume(int t) { 52 for (; top != t; --top) 53 if (s[top] < 0) --d[-s[top]]; 54 else fa[s[top]] = s[top], a[s[top]] = 0; 55 } 56 } 57 using namespace set; 58 59 void work(int l, int r, int m) { 60 int mid = l + r >> 1, now = top; 61 int i, j, fx, fy, _d; 62 for (i = 1; i <= m; ++i) 63 if (e[i].st <= l && r <= e[i].ed) { 64 fx = find(e[i].x), fy = find(e[i].y), _d = !(dep(e[i].x) ^ dep(e[i].y)); 65 if (fx != fy) set_union(fx, fy, _d); 66 else if (_d) { 67 while (l <= r) 68 puts("No"), ++l; 69 set_resume(now); 70 return; 71 } 72 swap(e[m--], e[i--]); 73 }; 74 if (l == r) puts("Yes"); 75 else { 76 for (i = 1, j = m; i <= j; ++i) 77 if (e[i].st > mid) swap(e[j--], e[i--]); 78 work(l, mid, j); 79 for (i = 1, j = m; i <= j; ++i) 80 if (e[i].ed <= mid) swap(e[j--], e[i--]); 81 work(mid + 1, r, j); 82 } 83 set_resume(now); 84 } 85 86 int main() { 87 int i; 88 n = read(), m = read(), time = read(); 89 for (i = 1; i <= n; ++i) 90 fa[i] = i, d[i] = 1, a[i] = 0; 91 for (i = 1; i <= m; ++i) e[i].get(); 92 for (i = 1; i <= m; ++i) 93 if (e[i].st > e[i].ed) swap(e[i--], e[m--]); 94 work(1, time, m); 95 return 0; 96 } 97 98 inline int read() { 99 static int x, sgn; 100 static char ch; 101 x = 0, sgn = 1, ch = getchar(); 102 while (ch < '0' || '9' < ch) { 103 if (ch == '-') sgn = -1; 104 ch = getchar(); 105 } 106 while ('0' <= ch && ch <= '9') { 107 x = x * 10 + ch - '0'; 108 ch = getchar(); 109 } 110 return sgn * x; 111 }