I - Tunnel Warfare(线段树，区间合并)

During the War of Resistance Against Japan, tunnel warfare was carried out extensively in the vast areas of north China Plain. Generally speaking, villages connected by tunnels lay in a line. Except the two at the ends, every village was directly connected with two neighboring ones.

Frequently the invaders launched attack on some of the villages and destroyed the parts of tunnels in them. The Eighth Route Army commanders requested the latest connection state of the tunnels and villages. If some villages are severely isolated, restoration of connection must be done immediately!
Input
The first line of the input contains two positive integers n and m (n, m ≤ 50,000) indicating the number of villages and events. Each of the next m lines describes an event.

There are three different events described in different format shown below:

D x: The x-th village was destroyed.

Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.

R: The village destroyed last was rebuilt.
Output
Output the answer to each of the Army commanders’ request in order on a separate line.
Sample Input

7 9
D 3
D 6
D 5
Q 4
Q 5
R
Q 4
R
Q 4

Sample Output

1
0
2
4

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int s[56575];
struct node
{
  int l, r;//左右区间边界
  int lb, rb, mb;//左子树最大区间，右子树最大区间，中间的最大区间
}a[556755];
void Build(int node, int l, int r)//建树
{
  //更新
  a[node].l = l;
  a[node].r = r;
  a[node].lb = a[node].rb = a[node].mb = r - l + 1;
  if(l!=r)
  {
      int m = (r + l) / 2;
      Build(node*2, l, m);
      Build(node*2+1, m+1, r);
  }
}
void UpData(int node, int t, int add)
{
   if(a[node].l==a[node].r)
   {
      if(add==1)//修复
      {
        a[node].lb = a[node].rb = a[node].mb = 1;
      }
      else//消除
      {
         a[node].lb = a[node].rb = a[node].mb = 0;
      }
      return ;
   }
   int m = (a[node].l + a[node].r) / 2;
   if(t<=m)//更新
   UpData(node*2, t, add);
   else
   UpData(node*2+1, t, add);
   a[node].lb = a[node*2].lb;
   a[node].rb = a[node*2+1].rb;
   a[node].mb = max(max(a[node*2].mb, a[node*2+1].mb), a[node*2].rb+a[node*2+1].lb);//取三者之间最大的
   if(a[node*2].lb==a[node*2].r-a[node*2].l+1)
   {
     a[node].lb += a[node*2+1].lb;
   }
   if(a[node*2+1].rb==a[node*2+1].r-a[node*2+1].l+1)
   {
     a[node].rb += a[node*2].rb;
   }
}
int Query(int node, int t)//求个数
{
if(a[node].l==a[node].r||a[node].mb==0||a[node].mb==a[node].r-a[node].l+1)
  {
    return a[node].mb;
  }
  int m = (a[node].l + a[node].r) / 2;
  if(t<=m)
  {
    if(t>=a[node*2].r-a[node*2].rb+1)
    {
       return Query(node*2, t)+Query(2*node+1, m+1);
    }
    else
    {
      return Query(node*2, t);
    }
  }
  else
  {
    if(t<=a[node*2+1].l+a[node*2+1].lb-1)
    {
      return Query(node*2+1, t)+Query(node*2, m);
    }
    else
    return Query(node*2+1, t);
  }
}
int main()
{
  int n, m;
  while(~scanf("%d %d", &n, &m))
  {
      int top = -1;
      Build(1, 1, n);
      while(m--)
      {
        int t;
        char p[12];
        scanf("%s", p);
        if(p[0]=='D')
        {
           scanf("%d", &t);
           s[++top] = t;
           UpData(1, t, 0);
        }
        else if(p[0]=='R')
        {
          if(top>=0)
          {
          UpData(1, s[top], 1);
          top--;
          }
        }
        else
        {
            scanf("%d", &t);
           printf("%d\n", Query(1, t));
        }
      }
  }
  return 0;
}