Tunnel Warfare平衡树

E - Tunnel Warfare

Time Limit:1000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

Submit Status

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:

1. D x: The x-th village was destroyed.
2. Q x: The Army commands requested the number of villages that x-th village was directly or indirectly connected with including itself.
3. 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

Hint

An illustration of the sample input:

      OOOOOOO 
D 3   OOXOOOO 
D 6   OOXOOXO 
D 5   OOXOXXO 
R     OOXOOXO 
R     OOXOOOO

 

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
using namespace std;
const int MAXN = 50005;
const int MOD = 2147483647;

struct Node
{
     int left;      
     int right;     
     int size;     
     int val;       
     int key;       
}tree[MAXN];

int root,tot;

int add(int val){
    
    tot++;
    tree[tot].size = 1;
    tree[tot].val = val;
    tree[tot].key = rand()%MOD;
    tree[tot].left = 0;
    tree[tot].right = 0;

    return tot;
}

void update_root(int now){
    int left,right;

    left = tree[now].left;
    right = tree[now].right;

    tree[now].size = tree[left].size + tree[right].size + 1;
}

void split_new(int now, int &a, int &b, int val){
    
    if(now == 0){
        a = 0;
        b = 0;
        return;
    }

    if(tree[now].val <= val){
        a = now;
        split_new(tree[now].right, tree[a].right, b, val);
    }else{
        b = now;
        split_new(tree[now].left, a, tree[b].left, val);
    }

    update_root(now);
}

void merge_new(int &now, int a, int b){
    
    if(a==0 || b==0){
        now = a+b;
        return;
    }
    if(tree[a].key<tree[b].key){
        
        now = a;
        merge_new(tree[now].right, tree[a].right, b);
         
    }else{
        now = b;
        merge_new(tree[now].left, a, tree[b].left);
    }

    update_root(now);
}

int find_new(int now, int rank){
    if(rank==0){
        return 0;
    }
    
    while (tree[tree[now].left].size+1 != rank){
        if(tree[tree[now].left].size >= rank){
            now = tree[now].left;
        }else{
            rank -= tree[tree[now].left].size + 1;
            now = tree[now].right;
        }
    }
    
    return tree[now].val;
}

void insert_new(int val){
    int x,y,z;

    x = 0;
    y = 0;
    z = add(val);
    split_new(root, x, y, val);
    merge_new(x,x,z);
    merge_new(root, x, y);
}

void del_new(int val){
    int x,y,z;

    x = y = z = 0;
    split_new(root, x, y, val);
    split_new(x, x, z, val-1);
    merge_new(z, tree[z].left, tree[z].right);
    merge_new(x, x, z);
    merge_new(root, x, y);
}

int  get_pre(int val){
    int x,y;

    x = y = 0;
    split_new(root, x, y, val-1);
  int u=find_new(x, tree[x].size);
    merge_new(root, x, y);
    return u;
}

int get_next(int val){
    int x,y;
    
    x = y = 0;
    
    split_new(root,x,y,val);
    int u=find_new(y,1);
    merge_new(root,x,y);
    
    return u;
}

int opp[MAXN],d[MAXN];

void solve(){
    
    char op[2];
    int x,y,n,m;
    memset(tree,0,sizeof(tree));
    add(MOD);
    root=1;
//    tree[root].size =0;
    int s=0;
    int j=1;
    
    while(~scanf("%d%d",&n,&m)){
    
    insert_new(0);    
    insert_new(n+1);
    
        for(int i=1;i<=m;i++){    
            cin>>op;
                    
            if(op[0]=='D'){
                cin>>x;
                d[x]=1;
                opp[++s]=x;     
                insert_new(x);
            }
            
            if(op[0]=='Q'){
                cin>>x;
                if(d[x]==1){
                    printf("0\n");
                    continue;
                }
                
                int u1=get_next(x);
                int u2=get_pre(x);

                printf("%d\n",u1-u2-1);
            }
            
            if(op[0]=='R'){
                int x=opp[s];
                int xx=opp[s--];
                del_new(x);
                d[xx]=0;    
            }
            
        }
        
    }
    
} 

int main(){
    solve();
    return 0;
}