poj 2892

Tunnel Warfare

Time Limit: 1000MS		Memory Limit: 131072K
Total Submissions: 7725		Accepted: 3188

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 nextm 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

Source

POJ Monthly--2006.07.30, updog

这题被用作学平衡树的模板题，用的是treap。将被删除的村庄存入treap，若修复则删除，求连续序列用find函数。

#include<ctime>
#include<cstdio>
#include<cstdlib>
#include<iostream>
using namespace std;
int n,m,root=0,d[50100],st,ed,sz;
struct Treap{int rnd,v,w,l,r;}tr[80100];
inline void lturn(int &k)
{
    int t=tr[k].r;
    tr[k].r=tr[t].l;
    tr[t].l=k;
    k=t;
}
inline void rturn(int &k)
{
    int t=tr[k].l;
    tr[k].l=tr[t].r;
    tr[t].r=k;
    k=t;
}
inline void insert(int &w,int x)
{
    if(w==0){
        sz++;
        w=sz;
        tr[w].v=x;
        tr[w].w=1;
        tr[w].rnd=rand();
        return;
    }
    if(tr[w].v==x){
        tr[w].w++;
        return;
    }
    if(tr[w].v>x){
        insert(tr[w].l,x);
        if(tr[tr[w].l].rnd<tr[w].rnd)rturn(w);
    }
    else{
        insert(tr[w].r,x);
        if(tr[tr[w].r].rnd<tr[w].rnd)lturn(w);
    }
}
inline void del(int &w,int x)
{
    if(tr[w].v==x){
        if(tr[w].w>1){tr[w].w--;return;}
        if(tr[w].l*tr[w].r==0)w=tr[w].l+tr[w].r;
        else{
            if(tr[tr[w].l].rnd<tr[tr[w].r].rnd){
                rturn(w);
                del(w,x);}
            else{
                lturn(w);
                del(w,x);}
        }
        return;
    }
    else if(tr[w].v<x)del(tr[w].r,x);
    else del(tr[w].l,x);
}
inline void find(int &w,int x)
{
    if(w==0)return;
    if(tr[w].v>=x&&tr[w].v<ed)ed=tr[w].v;
    if(tr[w].v<=x&&tr[w].v>st)st=tr[w].v;
    if(tr[w].v<x)find(tr[w].r,x);
    else find(tr[w].l,x);
}
int main()
{
    srand(time(0));
    scanf("%d%d",&n,&m);
    char od;
    int a;
    for(int i=1,j=0;i<=m;i++){
        cin>>od;
        if(od=='D'){
            scanf("%d",&a);
            insert(root,a);
            j++;
            d[j]=a;
        }
        if(od=='R'){
            del(root,d[j]);
            j--;
        }
        if(od=='Q'){
            scanf("%d",&a);
            st=0,ed=n+1;
            find(root,a);
            if(st==a&&ed==a)puts("0");
            else printf("%d
",ed-st-1);
        }
    }
    return 0;
}