[POI2007]MEG-Megalopolis

题目描述

Byteotia has been eventually touched by globalisation, and so has Byteasar the Postman, who once roamedthe country lanes amidst sleepy hamlets and who now dashes down the motorways. But it is those strolls inthe days of yore that he reminisces about with a touch of tenderness.

In the olden days Byteotian villages (numbered from to ) were connected by bidirectional dirt roadsin such a way, that one could reach the village number (called Bitburg) from any other village in exactlyone way. This unique route passed only through villages with number less or equal to that of the startingvillage. Furthermore, each road connected exactly two distinct villages without passing through any othervillage. The roads did not intersect outside the villages, but tunnels and viaducts were not unheard of.

Time passing by, successive roads were being transformed into motorways. Byteasar remembers distinctly, when each of the country roads so disappeared. Nowadays, there is not a single country lane left in Byteotia - all of them have been replaced with motorways, which connect the villages into Byteotian Megalopolis.

Byteasar recalls his trips with post to those villages. Each time he was beginning his journey with letters to some distinct village in Bitburg. He asks you to calculate, for each such journey (which took place in a specific moment of time and led from Bitburg to a specified village), how many country roads it led through.

TaskWrite a programme which:

reads from the standard input:

descriptions of roads that once connected Byteotian villages, sequence of events: Byteasar‘s trips and the moments when respective roads were transformed into motorways, for each trip, calculates how many country roads Byteasar has had to walk, writes the outcome to the standard output.

在经济全球化浪潮的影响下,习惯于漫步在清晨的乡间小路的邮递员Blue Mary也开始骑着摩托车传递邮件了。不过，她经常回忆起以前在乡间漫步的情景。昔日，乡下有依次编号为1..n的n个小村庄，某些村庄之间有一些双向的 土路。从每个村庄都恰好有一条路径到达村庄1（即比特堡）。并且，对于每个村庄，它到比特堡的路径恰好只经过编号比它的编号小的村庄。另外，对于所有道路 而言，它们都不在除村庄以外的其他地点相遇。在这个未开化的地方，从来没有过高架桥和地下铁道。随着时间的推移，越来越多的土路被改造成了公路。至 今，Blue Mary还清晰地记得最后一条土路被改造为公路的情景。现在，这里已经没有土路了——所有的路都成为了公路，而昔日的村庄已经变成了一个大都市。 Blue Mary想起了在改造期间她送信的经历。她从比特堡出发，需要去某个村庄，并且在两次送信经历的间隔期间,有某些土路被改造成了公路.现在Blue Mary需要你的帮助：计算出每次送信她需要走过的土路数目。（对于公路，她可以骑摩托车；而对于土路，她就只好推车了。）

输入输出格式

In the first line of the standard input there is a single integer (),denoting the number of villages in Byteotia. The following lines contain descriptions of the roads, in the form of two integers, ()separated by a single space, denoting the numbers of villages connected with a road. Inthe next line there is a single integer (),denoting the number of trips Byteasar has made.

The following lines contain descriptions of the events, in chronological order:

A description of the form "A "(for ) denotes a country road between villages and beingtransformed into a motorway in that particular moment.

A description of the from "W " denotes Byteasar‘s trip from Bitburg to village .

Your programme should write out exactly integers to the standard output, one a line, denoting the numberof country roads Byteasar has travelled during his successive trips.

输入输出样例

51 21 31 44 54W 5A 1 4W 5A 4 5W 5W 2A 1 2A 1 3

2101

转化为DFS序
DFS序一个很重要的特性是两个相同节点之间的节点编号是这个节点的子树
在第一个节点编号+1，最后一个节点编号-1，维护前缀和即可查询路径
注意输出答案时-1，因为节点1还记了一个1

#include <bits/stdc++.h>inline void read(int &x){x = 0;char ch = getchar();char c = ch;while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();while(ch >= ‘0‘ && ch <= ‘9‘)x = x * 10 + ch - ‘0‘, ch = getchar();} inline void read(long long &x){x = 0;char ch = getchar();char c = ch;while(ch < ‘0‘ || ch > ‘9‘)c = ch, ch = getchar();while(ch >= ‘0‘ && ch <= ‘9‘)x = x * 10 + ch - ‘0‘, ch = getchar();} inline void swap(int &a, int &b){int tmp = a;a = b;b = tmp;}inline void swap(long long &a, long long &b){long long tmp = a;a = b;b = tmp;}inline int max(int &a, int &b){return a > b ? a : b;}inline long long max(long long &a, long long &b){return a > b ? a : b;}inline int min(int &a, int &b){return a < b ? a : b;}inline long long min(long long &a, long long &b){return a < b ? a : b;}inline int lowbit(int a){return a & (-a);}inline long long lowbit(long long a){return a & (-a);}const int INF = 0x3f3f3f3f;const int MAXN = 500000 + 10;const int MAXM = 500000 + 10;struct Edge{int u,v,next;}edge[MAXN << 1];int head[MAXN],cnt,n,m,tmp1,tmp2;void insert(int a, int b){edge[++cnt] = Edge{a,b,head[a]};head[a] = cnt;}int stack[MAXN], b[MAXN], top;int bit[MAXN];int l[MAXN << 1], r[MAXN << 1], rank;//l[i] means the begining of i//r[i] means the end of iinline void modify(int p, int k){	for(;p <= (n << 1);p += lowbit(p))		bit[p] += k;}inline int ask(int p){	int ans = 0;	for(;p;p -= lowbit(p))		ans += bit[p];	return ans;}void dfs(){	stack[++top] = 1;	while(top)	{		int u = stack[top --];		b[u] = true;		if(l[u])		{			r[u] = ++rank;			continue;		}		stack[++top] = u;		l[u] = ++rank;		for(int pos = head[u];pos;pos = edge[pos].next)		{			int v = edge[pos].v;			if(b[v])continue;			stack[++top] = v;			b[v] = true;		}	}}int main(){	read(n);	for(int i = 1;i < n;++i)	{		read(tmp1);read(tmp2);		insert(tmp1, tmp2);		insert(tmp2, tmp1);	}		dfs();	for(int i = 1;i <= n;++i)	{		modify(l[i], 1);		modify(r[i], -1);	}	read(m);	m = n + m - 1; 	for(int i = 1;i <= m;i ++)	{		char c = getchar();		while(c != ‘A‘ && c != ‘W‘)c = getchar();		if(c == ‘A‘)		{			read(tmp1);read(tmp2);			modify(l[tmp2], -1);			modify(r[tmp2], 1);		}		else		{			read(tmp1);			printf("%d\n", ask(l[tmp1]) - 1);		}	}	return 0;}