[POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103-CFANZ编程社区

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 [POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_#define

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

输入格式：

In the first line of the standard input there is a single integer [POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_i++_02

The following [POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_#define_03

A description of the form "A [POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_#include_04

A description of the from "W [POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_i++_05

输出格式：

Your programme should write out exactly [POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_#include_06

输入输出样例

输入样例#1：

复制

输出样例#1： 复制

[POI2007]MEG-Megalopolis 树的dfs序+树状数组维护差分 BZOJ1103_#define_07

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#include
//#pragma GCC optimize(2)
using namespace std;
#define maxn 260005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
#define mclr(x,a) memset((x),a,sizeof(x))
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 98765431;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-5
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair pii;

inline int rd() {
  int x = 0;
  char c = getchar();
  bool f = false;
  while (!isdigit(c)) {
    if (c == '-') f = true;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return f ? -x : x;
}


ll gcd(ll a, ll b) {
  return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }



/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
  if (!b) {
    x = 1; y = 0; return a;
  }
  ans = exgcd(b, a%b, x, y);
  ll t = x; x = y; y = t - a / b * y;
  return ans;
}
*/

int n;
int IN[maxn], OUT[maxn];
int clk;
struct node {
  int u, v, nxt;
}e[maxn];
int tot;
int head[maxn];
void addedge(int u, int v) {
  e[++tot].u = u; e[tot].v = v; e[tot].nxt = head[u];
  head[u] = tot;
}

void dfs1(int u, int fa) {
  IN[u] = ++clk;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa)continue;
    dfs1(v, u);
  }
  OUT[u] = clk;
}

int c[maxn];
void add(int x, int val) {
  while (x < maxn) {
    c[x] += val; x += x & -x;
  }
}
int query(int x) {
  int ans = 0;
  while (x > 0) {
    ans += c[x]; x -= x & -x;
  }
  return ans;
}
int u[maxn], v[maxn];

int main()
{
  //  ios::sync_with_stdio(0);
  n = rd();
  for (int i = 1; i < n; i++) {
    u[i] = rd(); v[i] = rd();
    addedge(u[i], v[i]); addedge(v[i], u[i]);
  }
  dfs1(1, 0);
  for (int i = 1; i < n; i++) {
    if (IN[u[i]] < IN[v[i]])swap(u[i], v[i]);
    add(IN[u[i]], 1); add(OUT[u[i]] + 1, -1);
  }
  int m = rd();
  for (int i = 1; i <= n + m - 1; i++) {
    char opt[3]; rdstr(opt);
    if (opt[0] == 'A') {
      int a = rd(), b = rd();
      if (IN[a] < IN[b])swap(a, b);
      add(IN[a], -1); add(OUT[a] + 1, 1);
    }
    else {
      int a = rd();
      printf("%d\n", query(IN[a]));
    }
  }
  return 0;
}