Memory units are numbered from 1 up to N. 
A sequence of memory units is called a memory block. 
The memory control system we consider now has four kinds of operations: 
1.  Reset Reset all memory units free. 
2.  New x Allocate a memory block consisted of x continuous free memory units with the least start number 
3.  Free x Release the memory block which includes unit x 
4.  Get x Return the start number of the xth memory block(Note that we count the memory blocks allocated from left to right) 
Where 1<=x<=N.You are request to find out the output for M operations. 

6 10
New 2
New 5
New 2
New 2
Free 3
Get 1
Get 2
Get 3
Free 3
Reset

New at 1
Reject New
New at 3
New at 5 
Free from 3 to 4
Get at 1
Get at 5
Reject Get
Reject Free
Reset Now

区间维护依旧与Hotel题目类似。变得更加复杂的是对Free get操作时区间的维护。采用引入vector记录区间的办法，为了提高效率，每次查询、插入都需要采用二分法（利用了vector插入时是在
给定地址前插入数据的特性）这种二分法的使用值得复习反思。

  1 #include <iostream>
  2 //#include<bits/stdc++.h>
  3 #include <stack>
  4 #include <queue>
  5 #include <map>
  6 #include <set>
  7 #include <cstdio>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <math.h>
 11 using namespace std;
 12 int N,M;
 13 const int MAX=5e4+5;
 14 struct node
 15 {
 16     int left,right,mid;
 17     int al,ar;
 18     int status;
 19     int len()
 20     {
 21         return ar-al+1;
 22     }
 23     void actl()
 24     {
 25         left=right=mid=(status?0:len());
 26     }
 27 }st[10*MAX];
 28 struct seg//记录区间
 29 {
 30     int left,right;
 31 };
 32 vector <seg> ve;//储存区间信息
 33 seg oh;//临时储存
 34 void init(int l,int r,int k)
 35 {
 36     st[k].al=l;
 37     st[k].ar=r;
 38     st[k].status=0;
 39     st[k].actl();
 40     if(l==r)
 41         return;
 42     init(l,(l+r)/2,2*k);
 43     init((l+r)/2+1,r,2*k+1);
 44 }
 45 void pushdown(int k)
 46 {
 47     if(st[k].status!=-1)
 48     {
 49         st[2*k].status=st[2*k+1].status=st[k].status;
 50         st[k].status=-1;
 51         st[2*k].actl();
 52         st[2*k+1].actl();
 53     }
 54 }
 55 int query(int len,int k)
 56 {
 57     if(len==1&&st[k].al==st[k].ar)
 58         return st[k].al;
 59     if(st[k].mid<len)
 60         return -1;
 61     pushdown(k);
 62     if(st[2*k].mid>=len)
 63         return query(len,2*k);
 64     else if(st[2*k].right+st[2*k+1].left>=len)
 65         return st[2*k].ar-st[2*k].right+1;
 66     else if(st[2*k+1].mid>=len)
 67         return query(len,2*k+1);
 68     return -1;
 69 }
 70 void update(int ul,int ur,int l,int r,int k,int val)
 71 {
 72     if(l==ul&&r==ur)
 73     {
 74         st[k].status=val;
 75         st[k].actl();
 76         return;
 77     }
 78     if(l==r)
 79         return;
 80     pushdown(k);
 81     int mid=(l+r)/2;
 82     if(ur<=mid)
 83         update(ul,ur,l,mid,2*k,val);
 84     else if(ul>mid)
 85         update(ul,ur,mid+1,r,2*k+1,val);
 86     else
 87     {
 88         update(ul,mid,l,mid,2*k,val);
 89         update(mid+1,ur,mid+1,r,2*k+1,val);
 90     }
 91     st[k].mid=max(st[2*k].mid,st[2*k+1].mid);
 92     st[k].left=st[2*k].left;
 93     st[k].right=st[2*k+1].right;
 94     st[k].mid=max(st[2*k].right+st[2*k+1].left,st[k].mid);
 95     if(st[2*k].left==st[2*k].len())
 96         st[k].left+=st[2*k+1].left;
 97     if(st[2*k+1].right==st[2*k+1].len())
 98         st[k].right+=st[2*k].right;
 99     return;
100 }
101 char command[10];
102 int tem,an;
103 int lo;//添加的位置
104 int binarysearch(int k)
105 {
106     int l,r,mid;
107     l=0;r=ve.size()-1;
108     while(l<=r)
109     {
110         mid=(l+r)/2;
111         if(ve[mid].left<=k)
112             l=mid+1;
113         else if(ve[mid].left>k)
114             r=mid-1;
115     }
116     return l;
117 }
118 int main()
119 {
120     while(~scanf("%d %d",&N,&M))
121     {
122         ve.clear();
123         init(1,N,1);
124         while(M--)
125         {
126             scanf("%s",command);
127             if(command[0]==‘R‘)
128             {
129                 ve.clear();
130                 st[1].status=0;
131                 st[1].actl();
132                 pushdown(1);
133                 printf("Reset Now\n");
134             }
135             else if(command[0]==‘N‘)
136             {
137                 scanf("%d",&tem);
138                 an=query(tem,1);
139                 if(an==-1)
140                     printf("Reject New\n");
141                 else
142                 {
143                     oh.left=an;oh.right=an+tem-1;
144                     lo=binarysearch(an);
145                     ve.insert(ve.begin()+lo,oh);
146 
147                     update(an,an+tem-1,1,N,1,1);
148                     printf("New at %d\n",an);
149                 }
150             }
151             else if(command[0]==‘F‘)
152             {
153                 scanf("%d",&tem);
154                 lo=binarysearch(tem)-1;
155                 if(lo<=-1||ve[lo].right<tem)
156                     printf("Reject Free\n");
157                 else
158                 {
159                     update(ve[lo].left,ve[lo].right,1,N,1,0);
160                     printf("Free from %d to %d\n",ve[lo].left,ve[lo].right);
161                     ve.erase(ve.begin()+lo,ve.begin()+lo+1);//删除掉ve中的这个seg区间
162                 }
163             }
164             else if(command[0]==‘G‘)
165             {
166                 scanf("%d",&tem);
167                 if(tem>ve.size())
168                     printf("Reject Get\n");
169                 else
170                 {
171                     printf("Get at %d\n",ve[tem-1].left);
172                 }
173             }
174         }
175         printf("\n");//题意要求两组数据之间换行
176     }
177 }