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 rto 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, and D_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.

  1 /**  2  * poj  3  * Problem#3667  4  * Accepted  5  * Time:1735ms  6  * Memory:5388k  7  */  8 #include<iostream>  9 #include<fstream>  10 #include<sstream> 11 #include<algorithm> 12 #include<cstdio> 13 #include<cstring> 14 #include<cstdlib> 15 #include<cctype> 16 #include<cmath> 17 #include<ctime> 18 #include<map> 19 #include<stack> 20 #include<set> 21 #include<queue> 22 #include<vector> 23 #ifndef WIN32 24 #define AUTO "%lld" 25 #else 26 #define AUTO "%I64d" 27 #endif 28 using namespace std; 29 typedef bool boolean; 30 #define inf 0xfffffff 31 #define smin(a, b) (a) = min((a), (b)) 32 #define smax(a, b) (a) = max((a), (b)) 33 #define max3(a, b, c)    max((a), max((b), (c))) 34 template<typename T> 35 inline boolean readInteger(T& u) { 36     char x; 37     int aFlag = 1; 38     while(!isdigit((x = getchar())) && x != ‘-‘ && x != -1); 39     if(x == -1)    { 40         ungetc(x, stdin); 41         return false; 42     } 43     if(x == ‘-‘) { 44         aFlag = -1; 45         x = getchar(); 46     } 47     for(u = x - ‘0‘; isdigit((x = getchar())); u = u * 10 + x - ‘0‘); 48     u *= aFlag; 49     ungetc(x, stdin); 50     return true; 51 } 52  53 typedef class SegTreeNode { 54     public: 55         int len; 56         int llen; 57         int rlen; 58         int lazy; 59         SegTreeNode* left, *right; 60         SegTreeNode(const int len = 1, const int llen = 1, const int rlen = 1):len(len), llen(llen), rlen(rlen), lazy(0) { 61             left = right = NULL; 62         } 63          64         inline void pushUp(int l, int r, int mid) { 65             len = max3(left->len, right->len, left->rlen + right->llen); 66             llen = left->llen, rlen = right->rlen; 67             if(left->llen == (mid - l + 1))    llen = left->llen + right->llen; 68             if(right->rlen == (r - mid))    rlen = left->rlen + right->rlen;     69         } 70          71         inline void pushDown(int l, int r, int mid) { 72             this->left->len = this->left->llen = this->left->rlen = (lazy > 0) ? (mid - l + 1) : (0); 73             this->right->len= this->right->llen= this->right->rlen= (lazy > 0) ? (r - mid) : (0); 74             this->left->lazy = this->right->lazy = lazy; 75             lazy = 0; 76         } 77 }SegTreeNode; 78  79 typedef class SegTree { 80     public: 81         SegTreeNode *root; 82          83         SegTree():root(NULL) {         } 84         SegTree(int s) { 85             build(root, 1, s); 86         } 87          88         void build(SegTreeNode*& node, int l, int r) { 89             node = new SegTreeNode(); 90             if(l == r)        return; 91             int mid = (l + r) >> 1; 92             build(node->left, l, mid); 93             build(node->right, mid + 1, r); 94             node->pushUp(l, r, mid); 95         } 96          97         void update(SegTreeNode*& node, int l, int r, int cl, int cr, int val) { 98             int mid = (l + r) >> 1; 99             if(node->lazy && l != r)    node->pushDown(l, r, mid);100             if(cl == l && r == cr) {101                 node->lazy = (val) ? (1) : (-1);102                 node->len = node->llen = node->rlen = val * (r - l + 1);103                 return;104             }105             if(cr <= mid)    update(node->left, l, mid, cl, cr, val);106             else if(cl > mid)    update(node->right, mid + 1, r, cl, cr, val);107             else {108                 update(node->left, l, mid, cl, mid, val);109                 update(node->right, mid + 1, r, mid + 1, cr, val);110             }111             node->pushUp(l, r, mid);112         }113         114         int query(SegTreeNode*& node, int l, int r, int len) {115             int mid = (l + r) >> 1;116             if(node->lazy && l != r)    node->pushDown(l, r, mid);117             if(r - l + 1 == len)    return l;118             if(l == r)    return 0;119             if(node->left->len >= len)    return query(node->left, l, mid, len);120             else if(node->left->rlen + node->right->llen >= len && node->left->rlen > 0)121                 return mid - node->left->rlen + 1;122             else if(node->right->len >= len)    return query(node->right, mid + 1, r, len);123             return 0;124         }125         126 }SegTree;127 128 int n, m;129 SegTree st;130 131 inline void solve() {132     readInteger(n);133     readInteger(m);134     st = SegTree(n);135     for(int i = 1, a, b, c; i <= m; i++) {136         readInteger(a);137         readInteger(b);138         if(a == 1) {139             c = st.query(st.root, 1, n, b);140             if(c != 0)    st.update(st.root, 1, n, c, b + c - 1, 0);141             printf("%d\n", c);142         } else {143             readInteger(c);144             st.update(st.root, 1, n, b, b + c - 1, 1);145         }146     }147 }148 149 int main() {150     solve();151     return 0;152 }