首先我们离线，把操作倒过来，就变成每次把一棵树连到1所在的树上
我们可以通过并查集来完成，然后就是每次并到1所在的树上时dfs一下这棵子树就好了
1 /************************************************************** 2 Problem: 2610 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:760 ms 7 Memory:12528 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 const int N = 2e5 + 5; 15 16 struct edge { 17 int next, to; 18 19 edge() {} 20 edge(int _n, int _t) : next(_n), to(_t) {} 21 } e[N << 1]; 22 23 int n, m; 24 int son[N][2], del[N][2]; 25 int q[N << 1][2], fa[N], ans[N]; 26 int first[N], tot; 27 28 inline int read() { 29 int x = 0, sgn = 1; 30 char ch = getchar(); 31 while (ch < '0' || '9' < ch) { 32 if (ch == '-') sgn = -1; 33 ch = getchar(); 34 } 35 while ('0' <= ch && ch <= '9') { 36 x = x * 10 + ch - '0'; 37 ch = getchar(); 38 } 39 return sgn * x; 40 } 41 42 inline void Add_Edges(int x, int y) { 43 e[++tot] = edge(first[x], y), first[x] = tot; 44 e[++tot] = edge(first[y], x), first[y] = tot; 45 } 46 47 #define y e[x].to 48 void dfs(int p, int fa, int t) { 49 ans[p] = t; 50 int x; 51 for (x = first[p]; x; x = e[x].next) 52 if (y != fa) dfs(y, p, t); 53 } 54 #undef y 55 56 int find_fa(int x) { 57 return x == fa[x] ? x : fa[x] = find_fa(fa[x]); 58 } 59 60 inline void set_union(int x, int y, int t) { 61 if (find_fa(x) == find_fa(y)) return; 62 if (fa[x] == find_fa(1)) dfs(y, 0, t); 63 else if (fa[y] == find_fa(1)) dfs(x, 0, t); 64 else Add_Edges(x, y); 65 fa[fa[x]] = fa[y]; 66 } 67 68 int main() { 69 int i, j, x, y; 70 n = read(), m = read() - 1; 71 for (i = 1; i <= n; ++i) 72 son[i][0] = read(), son[i][1] = read(), fa[i] = i; 73 for (i = 0; i <= m; ++i) { 74 x = read(), y = read() - 1; 75 q[i][0] = x, q[i][1] = y, del[x][y] = 1; 76 } 77 for (i = 1; i <= n; ++i) 78 for (j = 0; j < 2; ++j) 79 if (!del[i][j] && son[i][j] > 0) set_union(i, son[i][j], -1); 80 for (i = m; ~i; --i) 81 set_union(q[i][0], son[q[i][0]][q[i][1]], i); 82 puts("-1"); 83 for (i = 2; i <= n; ++i) printf("%d ", ans[i]); 84 return 0; 85 }