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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#57808#4387. Static Query on TreeqinjianbinAC ✓250ms40284kbC++173.4kb2022-10-22 23:29:422022-10-22 23:29:44

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-10-22 23:29:44]
  • 评测
  • 测评结果:AC
  • 用时:250ms
  • 内存:40284kb
  • [2022-10-22 23:29:42]
  • 提交

answer

#include <bits/stdc++.h>
#define l(k) (k << 1)
#define r(k) (k << 1 | 1)
#define rep(i, a, b) for(int i = (a); i < (b); i++)
#define _for(i, a, b) for(int i = (a); i <= (b); i++)
using namespace std;

const int N = 2e5 + 10;
int ta[N << 2], tab[N << 2], tabc[N << 2];
int lazya[N << 2], lazyb[N << 2], lazyc[N << 2];
int d[N], fa[N], top[N], id[N], siz[N], son[N], cnt;
vector<int> g[N];
vector<pair<int, int>> A, B, C;
int n, q;

void dfs1(int u, int father)
{
	d[u] = d[father] + 1;
	fa[u] = father;
	siz[u] = 1;
	for(int v: g[u])
	{
		dfs1(v, u);
		siz[u] += siz[v];
		if(siz[son[u]] < siz[v]) son[u] = v;
	}
}

void dfs2(int u, int fa, int t)
{
	top[u] = t;
	id[u] = ++cnt;
	
	if(siz[u] == 1) return;
	dfs2(son[u], u, t);

	for(int v: g[u])
	{
		if(v == fa || v == son[u]) continue;
		dfs2(v, u, v);
	}
}

void add(int u, int v, int op)
{
	while(top[u] != top[v])
	{
		if(d[top[u]] < d[top[v]]) swap(u, v);
		if(!op) A.push_back({id[top[u]], id[u]});
		else B.push_back({id[top[u]], id[u]});
		u = fa[top[u]];
	}
	if(d[u] < d[v]) swap(u, v);
	if(!op) A.push_back({id[v], id[u]});
	else B.push_back({id[v], id[u]});
}

void updateA(int k, int l, int r)
{
	ta[k] = r - l + 1;
	lazya[k] = 1;
}

void updateB(int k, int l, int r)
{
	tab[k] = ta[k];
	lazyb[k] = 1;
}

void updateC(int k, int l, int r)
{
	tabc[k] = tab[k];
	lazyc[k] = 1;
}

void up(int k)
{
	ta[k] = ta[l(k)] + ta[r(k)];
	tab[k] = tab[l(k)] + tab[r(k)];
	tabc[k] = tabc[l(k)] + tabc[r(k)];
}

void down(int k, int l, int r)
{
	int m = l + r >> 1;
	if(lazya[k]) updateA(l(k), l, m), updateA(r(k), m + 1, r), lazya[k] = 0; 
	if(lazyb[k]) updateB(l(k), l, m), updateB(r(k), m + 1, r), lazyb[k] = 0; 
	if(lazyc[k]) updateC(l(k), l, m), updateC(r(k), m + 1, r), lazyc[k] = 0; 
}

void change(int k, int l, int r, int L, int R, int op)
{
	if(L <= l && r <= R)
	{
		if(!op) updateA(k, l, r);
		else if(op == 1) updateB(k, l, r);
		else updateC(k, l, r);
		return;
	}
	down(k, l, r);
	int m = l + r >> 1;
	if(L <= m) change(l(k), l, m, L, R, op);
	if(R > m) change(r(k), m + 1, r, L, R, op);
	up(k);
}

void cla(int k)
{
	ta[k] = tab[k] = tabc[k] = lazya[k] = lazyb[k] = lazyc[k] = 0;
}

void Clear(int k, int l, int r, int L, int R)
{
	if(L <= l && r <= R)
	{
		cla(k);
		return;
	}
	cla(l(k)); cla(r(k));
	int m = l + r >> 1;
	if(L <= m) Clear(l(k), l, m, L, R);
	if(R > m) Clear(r(k), m + 1, r, L, R);
}
 
int main()
{
	int T; scanf("%d", &T);
	while(T--)
	{
		scanf("%d%d", &n, &q);
		_for(i, 1, n) g[i].clear(), son[i] = d[i] = top[i] = fa[i] = id[i] = siz[i] = 0;
		cnt = 0;
		_for(i, 2, n)
		{
			int x; scanf("%d", &x);
			g[x].push_back(i);
		}

		dfs1(1, 0);
		dfs2(1, 0, 1);

		while(q--)
		{
			A.clear(); B.clear(); C.clear();

			int na, nb, nc, x;
			scanf("%d%d%d", &na, &nb, &nc);
			while(na--)
			{
				scanf("%d", &x);
				add(1, x, 0);
			}
			while(nb--)
			{
				scanf("%d", &x);
				add(1, x, 1);
			}
			while(nc--)
			{
				scanf("%d", &x);
				C.push_back({id[x], id[x] + siz[x] - 1});
			}

			for(auto [l, r]: A) change(1, 1, n, l, r, 0);
			for(auto [l, r]: B) change(1, 1, n, l, r, 1);
			for(auto [l, r]: C) change(1, 1, n, l, r, 2);
			printf("%d\n", tabc[1]);

			for(auto [l, r]: A) Clear(1, 1, n, l, r);
			for(auto [l, r]: B) Clear(1, 1, n, l, r);
			for(auto [l, r]: C) Clear(1, 1, n, l, r);
		}
	}

    return 0;
}

詳細信息

Test #1:

score: 100
Accepted
time: 250ms
memory: 40284kb

input:

1
200000 18309
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 ...

output:

102147
62590
87270
88880
7654
61542
62953
85022
55135
54125
70500
64356
25824
88300
42278
15336
18132
28734
90282
42889
28099
31311
96842
19959
34366
60205
78358
91142
56048
74688
86091
51979
94750
11989
89544
86860
56720
29534
52343
90031
79002
90293
94554
48340
65015
9181
15016
19884
49445
14181
6...

result:

ok 18309 numbers