HDU1540(线段树)

Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3047    Accepted Submission(s): 1162

Problem Description

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    //转载大神的，然后加上自己的理解，不知怎么就错了！
   
   
   
   
   
   
   
   

    
    
    
    
#include"stdio.h" #include"string.h" #include"stdlib.h" #include"stack" using namespace std; #define N 50011
    
    
    
    
int n,m; int hash[N]; struct Segtree {  int l,r,mid;  int count; }T[4*N];
    
    
    
    
void build(int l,int r,int k) {  T[k].l=l;  T[k].r=r;  T[k].mid=(l+r)>>1;  //叶子节点  if(l==r)  {   T[k].count=0;   return ;  }  build(l,T[k].mid,2*k);  build(T[k].mid+1,r,2*k+1);  T[k].count=0; }
    
    
    
    
int findl(int l,int r,int k) {  int ans=0;  //左右区间正好等于该节点左右区间  if(T[k].l==l && T[k].r==r && !T[k].count)   return r-l+1;  //结点为叶子节点  if(T[k].l==T[k].r && T[k].l==l && l==r)    return 1-T[k].count;  //位于区间左边  if(r<=T[k].mid)    ans=findl(l,r,2*k);  //位于区间左边  else if(l>T[k].mid)    ans=findl(l,r,2*k+1);  else  {   ans=findl(T[k].mid+1,r,2*k+1);//到区间右边搜索   if(ans==r-T[k].mid)///证明该区间为断开，需到中点左边搜索    ans+=findl(l,T[k].mid,2*k);  }  return ans; }
    
    
    
    
int findr(int l,int r,int k) {  int ans=0;  if(T[k].l==l && T[k].r==r && !T[k].count)   return r-l+1;  if(T[k].l==T[k].r && T[k].l==l && l==r)    return 1-T[k].count;  if(r<=T[k].mid)    ans=findr(l,r,2*k);  else if(l>T[k].mid)    ans=findr(l,r,2*k+1);  else  {   ans=findr(l,T[k].mid,2*k);   if(ans==T[k].mid-l+1)     ans+=findr(T[k].mid+1,r,2*k+1);  }  return ans; }
    
    
    
    
void Dir(int aim,int dir,int k) {  if(T[k].l==T[k].r && T[k].l==aim)  {   T[k].count+=dir;   return ;  }  if(aim<=T[k].mid)    Dir(aim,dir,2*k);  else      Dir(aim,dir,2*k+1);  T[k].count+=dir;     //这里不判0了，所以千万要先判一下hash }
    
    
    
    
int main() {  int temp;  char str[111];  while(scanf("%d%d",&n,&m)!=-1)  {   stack<int>st;   build(1,n,1);   memset(hash,0,sizeof(hash));   while(m--)   {    scanf("%s",str);    if(str[0]!='R')    {     scanf("%d",&temp);     if(str[0]=='D')     {      st.push(temp);      if(hash[temp])        continue;  //注意，hash亮了的话，说明被del了      else      {       hash[temp]=1;       Dir(temp,1,1);      }     }     else if(str[0]=='Q')     {      if(hash[temp])        printf("0\n");      else      {       //用左右分别查找吧       int a,b;       a=findl(1,temp,1);       b=findr(temp,n,1);       printf("%d\n",a+b-1);      }     }    }    else    {     if(!st.empty())     {      temp=st.top();      st.pop();      if(!hash[temp])       continue;      hash[temp]=0;      Dir(temp,-1,1);     }    }   }  }  return 0; }