3351: [ioi2009]Regions

3351: [ioi2009]Regions

Time Limit: 120 Sec   Memory Limit: 128 MB
Submit: 205   Solved: 59
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Description

 

N个节点的树，有R种属性，每个点属于一种属性。有Q次询问，每次询问r1,r2，回答有多少对(e1,e2)满足e1属性是r1，e2属性是r2，e1是e2的祖先。

数据规模
N≤200000,R≤25000，Q≤200000

30%数据R≤500

55%数据同种属性节点个数≤500

Input

Output

Sample Input

6 3 4
1
1 2
1 3
2 3
2 3
5 1
1 2
1 3
2 3
3 1

Sample Output

1
3
2
1

HINT

Source

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参考于《根号算法——不只是分块》 王悦同

设属性r1的点有A个属性r2的点有B个

若A很小，则设计一个AlogB的算法，若B很小，则设计一个BlogA的算法

若都很大，这样的询问不会很多，就直接O(N)暴力

针对不同类型询问设计不同的专杀算法，运用根号卡时间

苟蒻编写能力还有待提高！注意数组下标。。。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;

const int maxn = 2E5 + 20;

struct Q{
	int r1,r2,num;
	bool operator < (const Q &b) const {
		if (r1 < b.r1) return 1;
		if (r1 > b.r1) return 0;
		return r2 < b.r2;
	}
	bool operator == (const Q &b) const {
		return (r1 == b.r1 && r2 == b.r2);
	}
}q[maxn];

struct P{
	int pos,add,sum;
	bool operator < (const P &b) const {
		return pos < b.pos;
	}
};

vector <int> v[maxn],v2[maxn];
vector <P> v3[maxn];

int mark[maxn],dfs[maxn],n,m,ans[maxn],dfs_clock = 0,flag;
int head[maxn],tail[maxn];

int getint()
{
	int ret = 0;
	char ch = getchar();
	while (ch < '0' || ch > '9') ch = getchar();
	while ('0' <= ch && ch <= '9') {
		ret = ret*10 + ch - '0';
		ch = getchar();
	}
	return ret;
}

void DFS(int k)
{
	dfs[k] = head[k] = ++dfs_clock;
	v2[mark[k]].push_back(k);
	for (int i = 0; i < v[k].size(); i++) DFS(v[k][i]);
	tail[k] = dfs_clock;
	if (head[k] != tail[k]) {
		v3[mark[k]].push_back((P){head[k],1,1});
		v3[mark[k]].push_back((P){tail[k]+1,-1,-1});
	}
}

int DBSC1(int x,int l,int r,int pos)
{
	if (r - l <= 1) {
		if (dfs[v2[x][l]] >= pos) return l;
		else return r;
	}
	int mid = (l+r) >> 1;
	if (dfs[v2[x][mid]] >= pos) return DBSC1(x,l,mid,pos);
	else return DBSC1(x,mid,r,pos);
}

int DBSC2(int x,int l,int r,int pos)
{
	if (r - l <= 1) {
		if (dfs[v2[x][r]] <= pos) return r;
		else return l;
	} 
	int mid = (l+r) >> 1;
	if (dfs[v2[x][mid]] <= pos) return DBSC2(x,mid,r,pos);
	else return DBSC2(x,l,mid,pos);
}

int DBSC(int x,int l,int r,int pos)
{
	if (r - l <= 1) {
		if (v3[x][r].pos <= pos) return r;
		else return l;
	} 
	int mid = (l+r) >> 1;
	if (v3[x][mid].pos <= pos) return DBSC(x,mid,r,pos);
	else return DBSC(x,l,mid,pos);
}

void solve1(int x)
{
	int ANS = 0;
	int r1 = q[x].r1;
	int r2 = q[x].r2;
	for (int i = 0; i < v2[r1].size(); i++) {
		int now = v2[r1][i];
		if (head[now] == tail[now]) continue;
		if (tail[now] < dfs[v2[r2][0]]) continue;
		if (head[now] > dfs[v2[r2][v2[r2].size()-1]]) continue;
		int L = DBSC1(r2,0,v2[r2].size()-1,head[now]);
		int R = DBSC2(r2,0,v2[r2].size()-1,tail[now]);
		ANS += R-L+1;
	}
	ans[q[x].num] = ANS;
}

void solve2(int x)
{
	int ANS = 0;
	int r1 = q[x].r1;
	int r2 = q[x].r2;
	for (int i = 0; i < v2[r2].size(); i++) {
		int now = v2[r2][i];
		if (dfs[now] < v3[r1][0].pos) continue;
		if (dfs[now] > v3[r1][v3[r1].size()-1].pos) continue;
		/*for (int i = 0; i < v3[r1].size(); i++) {
			int pos = v3[r1][i].pos;
			int add = v3[r1][i].add;
			int sum = v3[r1][i].sum;
			int bbbb = 1;
		}*/
		int POS = DBSC(r1,0,v3[r1].size()-1,dfs[now]);
		ANS += v3[r1][POS].sum;
	}
	ans[q[x].num] = ANS;
}

void solve3(int x)
{
	int ANS,SUM,L,R;
	int r1 = q[x].r1;
	int r2 = q[x].r2;
	ANS = SUM = L = R = 0;
	
	while (L < v3[r1].size() && R < v2[r2].size()) {
		if (v3[r1][L].pos <= dfs[v2[r2][R]]) {
			SUM += v3[r1][L].add;
			L++; 
		}
		else {
			ANS += SUM;
			R++;
		}
	}
	
	//while (R < v2[r2].size()) ANS += SUM,R++;
	ans[q[x].num] = ANS;
}

int main()
{
	#ifdef YZY
		   freopen("yzy.txt","r",stdin);
	#endif
	
	int tt;
	cin >> n >> tt >> m >> mark[1];
	flag = sqrt(n);
	
	for (int i = 2; i <= n; i++) {
		int x,y;
		x = getint(); y = getint();
		v[x].push_back(i); mark[i] = y;
	}
	DFS(1);
	
	/*for (int i = 1; i <= tt; i++) {
		for (int j = 0; j < v2[i].size(); j++) {
			int kk = v2[i][j];
			int b = 1;
		}
		for (int j = 0; j < v3[i].size(); j++) {
			int pos = v3[i][j].pos;
			int add = v3[i][j].add;
			int sum = v3[i][j].sum;
			int b = 1;
		}
	}*/
	
	for (int i = 1; i <= tt; i++) {
		sort(v3[i].begin(),v3[i].end());
		for (int j = 1; j < v3[i].size(); j++) 
			v3[i][j].sum += v3[i][j-1].sum;
	}
	
	for (int i = 1; i <= m; i++) {
		int x,y;
		x = getint(); y = getint();
		q[i] = (Q){x,y,i};
	}
	sort(q+1,q+m+1);
	
	for (int i = 1; i <= m; i++) {
		if (q[i] == q[i-1]) {
			ans[q[i].num] = ans[q[i-1].num];
			continue;
		}
		int S1 = v2[q[i].r1].size();
		int S2 = v2[q[i].r2].size();
		if (S1 <= flag && S2 <= flag) {
			solve3(i);
			continue;
		}
		if (S1 <= flag) {
			solve1(i);
			continue;
		}
		if (S2 <= flag) {
			solve2(i);
			continue;
		}
		solve3(i);
	}
	
	for (int i = 1; i <= m; i++) printf("%d\n",ans[i]);
	
	return 0;
}