[CF620E]New Year Tree_dfs序_线段树_bitset
New Year Tree
题目链接:http://codeforces.com/problemset/problem/620/E
数据范围:略。
题解:
转化成序列问题,发现颜色种数特别少,暴力用数组合并显然会$T$,我们用$bitset$优化合并过程即可。
代码:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 | #include <bits/stdc++.h> #define setIO(s) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout) #define N 800010 #define ls p << 1 #define rs p << 1 | 1 using namespace std; char *p1, *p2, buf[100000]; #define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1 ++ ) int rd() { int x = 0, f = 1; char c = nc(); while (c < 48) { if (c == '-' ) f = -1; c = nc(); } while (c > 47) { x = (((x << 2) + x) << 1) + (c ^ 48), c = nc(); } return x * f; } int head[N], to[N << 1], nxt[N << 1], tot; inline void add( int x, int y) { to[ ++ tot] = y; nxt[tot] = head[x]; head[x] = tot; } bitset <60> a[N << 2], tag[N << 2], val, ans; inline void pushup( int p) { a[p] = a[ls] | a[rs]; } inline void pushdown( int p) { if (tag[p].count()) { a[ls] = tag[ls] = tag[p]; a[rs] = tag[rs] = tag[p]; tag[p].reset(); } } void update( int x, int y, int l, int r, int p) { if (x <= l && r <= y) { a[p] = tag[p] = val; return ; } int mid = (l + r) >> 1; pushdown(p); if (x <= mid) { update(x, y, l, mid, ls); } if (mid < y) { update(x, y, mid + 1, r, rs); } pushup(p); } void query( int x, int y, int l, int r, int p) { if (x <= l && r <= y) { ans = ans | a[p]; return ; } int mid = (l + r) >> 1; pushdown(p); if (x <= mid) { query(x, y, l, mid, ls); } if (mid < y) { query(x, y, mid + 1, r, rs); } } int sz[N], dic[N], cnt, re[N]; void dfs( int p, int fa) { dic[p] = ++cnt, re[cnt] = p; sz[p] = 1; for ( int i = head[p]; i; i = nxt[i]) { if (to[i] != fa) { dfs(to[i], p); sz[p] += sz[to[i]]; } } } int v[N]; void build( int l, int r, int p) { if (l == r) { a[p].set(v[re[l]] - 1); return ; } int mid = (l + r) >> 1; build(l, mid, ls); build(mid + 1, r, rs); pushup(p); } int main() { // setIO("data-structure"); int n = rd(), m = rd(); for ( int i = 1; i <= n; i ++ ) { v[i] = rd(); } for ( int i = 1; i < n; i ++ ) { int x = rd(), y = rd(); add(x, y), add(y, x); } dfs(1, 1); build(1, n, 1); for ( int i = 1; i <= m; i ++ ) { int opt = rd(); if (opt == 1) { int x = rd(), y = rd(); val.reset(); val.set(y - 1); update(dic[x], dic[x] + sz[x] - 1, 1, n, 1); } else { int x = rd(); ans.reset(); query(dic[x], dic[x] + sz[x] - 1, 1, n, 1); printf ( "%d\n" , ans.count()); } } fclose (stdin), fclose (stdout); return 0; } |
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