HDＵ－１５４０ Tunnel Warfare

　　　　　　　　　　Tunnel Warfare

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3419    Accepted Submission(s): 1328

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

同学的代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <stdlib.h>
using namespace std;

const int maxn=50000+10;

int n,m;
int s[maxn],top;

struct node
{
    int l,r;
    int ls,rs,ms;
}a[maxn<<2];

void init(int l,int r,int i)
{
    a[i].l = l;
    a[i].r = r;
    a[i].ls = a[i].rs = a[i].ms = r-l+1;
    if(l!=r)
    {
        int mid = (l+r)>>1;
        init(l,mid,i*2);
        init(mid+1,r,2*i+1);
    }
}
void insert(int i,int t,int x)
{
    if(a[i].l == a[i].r)
    {
        if(x==1)
        a[i].ls = a[i].rs = a[i].ms = 1;
        else
        a[i].ls=a[i].rs=a[i].ms=0;
        return ;
    }
    int mid =(a[i].l+a[i].r)>>1;
    if(t<=mid)
      insert(2*i,t,x);
    else
    insert(2*i+1,t,x);
    a[i].ls = a[2*i].ls;
    a[i].rs = a[2*i+1].rs;
    a[i].ms = max(max(a[2*i].ms,a[2*i+1].ms),a[2*i].rs+a[2*i+1].ls);
    if(a[2*i].ls == a[2*i].r-a[2*i].l+1)
    a[i].ls += a[2*i+1].ls;
    if(a[2*i+1].rs == a[2*i+1].r-a[2*i+1].l+1)
    a[i].rs += a[2*i].rs;
}

int query(int i,int t)
{
    if(a[i].l == a[i].r || a[i].ms == 0 || a[i].ms == a[i].r-a[i].l+1)
    return a[i].ms;
    int mid = (a[i].l+a[i].r)>>1;
    if(t<=mid)
    {
        if(t>=a[2*i].r-a[2*i].rs+1)
        return query(2*i,t)+query(2*i+1,mid+1);
        else
        return query(2*i,t);
    }
    else
    {
        if(t<=a[2*i+1].l+a[2*i+1].ls-1)
        return query(2*i+1,t)+query(2*i,mid);
        else
        return query(2*i+1,t);
    }
}

int main()
{
    int i,j,x;
    char ch[2];
    while(~scanf("%d%d",&n,&m))
    {
        top = 0;
        init(1,n,1);
        while(m--)
        {
            scanf("%s",ch);
            if(ch[0] == 'D')
            {
                scanf("%d",&x);
                s[top++] = x;
                insert(1,x,0);
            }
            else if(ch[0]=='Q')
            {
                scanf("%d",&x);
                printf("%d\n",query(1,x));
            }
            else
            {
                if(x>0)
                {
                    x = s[--top];
                    insert(1,x,1);
                }

           }
        }
    }
   return 0;
}