Hotel

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 2993		Accepted: 1143

The cows are journeying north to Thunder Bay in Canada to gain cultural enrichment and enjoy a vacation on the sunny shores of Lake Superior. Bessie, ever the competent travel agent, has named the Bullmoose Hotel on famed Cumberland Street as their vacation residence. This immense hotel has N (1 ≤ N ≤ 50,000) rooms all located on the same side of an extremely long hallway (all the better to see the lake, of course).

The cows and other visitors arrive in groups of size D_i (1 ≤ D_i ≤ N) and approach the front desk to check in. Each group i requests a set of D_i contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+D_i-1 if they are available or, if no contiguous set of rooms is available, politely suggests alternate lodging. Canmuu always chooses the value of r to be the smallest possible.

Visitors also depart the hotel from groups of contiguous rooms. Checkout i has the parameters X_i and D_i which specify the vacating of rooms X_i ..X_i +D_i-1 (1 ≤ X_i ≤ N-D_i+1). Some (or all) of those rooms might be empty before the checkout.

Your job is to assist Canmuu by processing M (1 ≤ M < 50,000) checkin/checkout requests. The hotel is initially unoccupied.

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Line i+1 contains request expressed as one of two possible formats: (a) Two space separated integers representing a check-in request: 1 and D_i (b) Three space-separated integers representing a check-out: 2, X_i, andD_i

* Lines 1.....: For each check-in request, output a single line with a single integer r, the first room in the contiguous sequence of rooms to be occupied. If the request cannot be satisfied, output 0.

10 6
1 3
1 3
1 3
1 3
2 5 5
1 6

1
4
7
0
5

/*
Mail to   : miyubai@gamil.com
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : PKU_3667
Doc Name  : HOTEL
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
inline int max ( int a, int b ) {
       return a > b ? a : b;       
}
struct segTree {
       int left, right, lVal, rVal, mVal, cov;// cov -- >  0: 线段为空   1: 线段为满  -1：2种情况都有 
       int mid () { return (left+right)>>1; }
       int dis () { return right - left + 1; }
       void set ( int flag ) { // 0: 线段为空   1: 线段为满 
            if ( flag ){
                 cov = 1;
                 lVal = rVal = mVal = 0;  
            } else {
                 cov = 0;
                 lVal = rVal = mVal = right - left + 1;     
            }
       }     
}seg[160000];
void creat ( int left, int right, int rt = 1 ) {   // 初始化线段 
     seg[rt].left = left;
     seg[rt].right = right;
     seg[rt].set (0);
     if ( left == right ) return;
     int LL = rt << 1, RR = rt << 1 | 1, mid = seg[rt].mid();  
     creat ( left, mid, LL );
     creat ( mid + 1, right, RR );  
}
void modify ( int left, int right, int flag, int rt = 1 ) {
     if ( seg[rt].left >= left && seg[rt].right <= right ) {   //如果线段被覆盖,  直接按flag标记处理,返回 
         seg[rt].set ( flag );   return;
     }     
     int LL = rt << 1, RR = rt << 1 | 1, mid = seg[rt].mid();
     if ( seg[rt].cov != -1 ) {     // 如果线段不是混合情况(即线段是被覆盖的), 把标志下传 
          seg[LL].cov = seg[RR].cov = seg[rt].cov;
          seg[LL].set ( seg[LL].cov );   
          seg[RR].set ( seg[RR].cov );
     } 
     if ( right <= mid ) modify ( left, right, flag, LL );    //递归更新线段 
     else if ( left > mid ) modify ( left, right, flag, RR );
     else {
           modify ( left, mid, flag, LL );
           modify ( mid + 1, right, flag, RR );     
     }
     if ( seg[LL].cov == 0 && seg[RR].cov == 0 ) seg[rt].cov = 0; //线段为空 
     else if ( seg[LL].cov == 1 && seg[RR].cov == 1 ) seg[rt].cov = 1; //线段满 
     else seg[rt].cov = -1;  // 2种情况都有 
     seg[rt].mVal = max(seg[LL].rVal+seg[RR].lVal,max(seg[LL].mVal, seg[RR].mVal)); // 线段的更新,  经典部分
     seg[rt].lVal = seg[LL].lVal + ( seg[LL].cov == 0 ? seg[RR].lVal : 0 );
     seg[rt].rVal = seg[RR].rVal + ( seg[RR].cov == 0 ? seg[LL].rVal : 0 );
}
int query ( int val, int rt = 1 ) {
    int LL = rt << 1, RR = rt << 1 | 1, mid = seg[rt].mid();
    if ( seg[rt].cov == 0 && seg[rt].mVal >= val ) {   //线段为空,且可用,直接返回线段左端点 
             return seg[rt].left;
    } else if ( seg[rt].cov == -1 ) {   //分三种 情况处理  左   左右    右  处理   
         if ( seg[LL].mVal >= val ) return query ( val, LL );
         else if ( seg[LL].rVal + seg[RR].lVal >= val ) return mid - seg[LL].rVal + 1;
         else if ( seg[RR].mVal >= val )return query ( val, RR );  
    }   
    return 0;
}
int main ()
{
    int N, M, left, right, val, choice;
    while ( scanf ( "%d%d", &N,&M ) == 2 ) {
           creat ( 1, N );
           while ( M -- ) {
                  scanf ( "%d", &choice );
                  switch ( choice ) {
                          case 1 : scanf ( "%d",&val );
                                   printf ( "%d\n", left = query ( val ) );
                                   if ( left != 0 ) {
                                        right = left + val - 1;
                                        modify ( left, right, 1 );
                                   }
                                   break;
                          case 2 : scanf ( "%d%d",&left, &val );
                                   right = left + val - 1;
                                   modify ( left, right, 0 );
                                   break;       
                  }       
           }      
    }
    return 0;
}