| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 18195 | Accepted: 7897 |
Description
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 Di (1 ≤ Di ≤ N) and approach the front desk to check in. Each group i requests a set of Di contiguous rooms from Canmuu, the moose staffing the counter. He assigns them some set of consecutive room numbers r..r+Di-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 Xi and Di which specify the vacating of rooms Xi ..Xi +Di-1 (1 ≤ Xi ≤ N-Di+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.
Input
* 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 Di (b) Three space-separated integers representing a check-out: 2, Xi, and Di
Output
* 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.
Sample Input
10 6 1 3 1 3 1 3 1 3 2 5 5 1 6
Sample Output
1 4 7 0 5
Source
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #define max(a, b) ((a) > (b) ? (a) :(b)) 6 #define min(a, b) ((a) < (b) ? (a) : (b)) 7 8 inline void read(int &x) 9 { 10 x = 0;char ch = getchar(), c = ch; 11 while(ch < '0' || ch > '9')c = ch, ch = getchar(); 12 while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar(); 13 if(c == '-')x = -x; 14 } 15 16 inline void swap(int &a, int &b) 17 { 18 int tmp = a;a = b;b = tmp; 19 } 20 21 const int MAXN = 100000 + 10; 22 23 int n,m,left[MAXN << 2], right[MAXN << 2], ma[MAXN << 2], sum[MAXN << 2], lazy[MAXN << 2]; 24 25 void build(int o, int l, int r) 26 { 27 lazy[o] = -1; 28 if(l == r) 29 { 30 left[o] = right[o] = ma[o] = sum[o] = 1; 31 return; 32 } 33 int mid = (l + r) >> 1; 34 build(o << 1, l, mid); 35 build(o << 1 | 1, mid + 1, r); 36 37 left[o] = left[o << 1]; 38 if(left[o << 1] == sum[o << 1])left[o] += left[o << 1 | 1]; 39 right[o] = right[o << 1 | 1]; 40 if(right[o << 1 | 1] == sum[o << 1 | 1])right[o] += right[o << 1]; 41 ma[o] = max(left[o], max(left[o << 1 | 1] + right[o << 1],right[o])); 42 ma[o] = max(ma[o], max(ma[o << 1], ma[o << 1 | 1])); 43 sum[o] = sum[o << 1] + sum[o << 1 | 1]; 44 } 45 46 inline void pushdown(int o, int l, int r) 47 { 48 if(lazy[o] == -1)return; 49 register int mid = (l + r) >> 1; 50 register int k = lazy[o]; 51 lazy[o << 1] = lazy[o << 1 | 1] = k; 52 left[o << 1] = right[o << 1] = ma[o << 1] = k * (mid - l + 1); 53 left[o << 1 | 1] = right[o << 1 | 1] = ma[o << 1 | 1] = k * (r - mid); 54 lazy[o] = -1; 55 } 56 57 void modify(int ll, int rr, int k, int o, int l, int r) 58 { 59 pushdown(o, l, r); 60 if(ll <= l && rr >= r) 61 { 62 left[o] = right[o] = ma[o] = k * (r - l + 1); 63 lazy[o] = k; 64 return; 65 } 66 int mid = (l + r) >> 1; 67 if(mid >= ll)modify(ll, rr, k, o << 1, l, mid); 68 if(mid < rr) modify(ll, rr, k, o << 1 | 1, mid + 1, r); 69 70 left[o] = left[o << 1]; 71 if(left[o << 1] == sum[o << 1])left[o] += left[o << 1 | 1]; 72 right[o] = right[o << 1 | 1]; 73 if(right[o << 1 | 1] == sum[o << 1 | 1])right[o] += right[o << 1]; 74 ma[o] = max(left[o], max(left[o << 1 | 1] + right[o << 1],right[o])); 75 ma[o] = max(ma[o], max(ma[o << 1], ma[o << 1 | 1])); 76 return; 77 } 78 79 struct Node 80 { 81 int left, sum; 82 Node(int _left, int _sum){left = _left; sum = _sum;} 83 Node(){left = sum = 0;} 84 }; 85 86 int ask(int p, int o, int l, int r) 87 { 88 pushdown(o, l, r); 89 if(l == r && left[l])return l; 90 int mid = (l + r) >> 1; 91 if(ma[o << 1] >= p)return ask(p, o << 1, l, mid); 92 else 93 { 94 if(right[o << 1] + left[o << 1 | 1] >= p) return mid - right[o << 1] + 1; 95 else if(ma[o << 1 | 1] >= p)return ask(p, o << 1 | 1, mid + 1, r); 96 else return 0; 97 } 98 } 99 100 int main() 101 { 102 read(n), read(m); 103 register int tmp1, tmp2, tmp3; 104 memset(lazy, -1, sizeof(lazy)); 105 build(1,1,n); 106 for(register int i = 1;i <= m;++ i) 107 { 108 read(tmp1); 109 if(tmp1 == 1) 110 { 111 read(tmp2); 112 tmp3 = ask(tmp2, 1, 1, n); 113 printf("%d ", tmp3); 114 if(tmp3)modify(tmp3, tmp3 + tmp2 - 1, 0, 1, 1, n); 115 } 116 else 117 { 118 read(tmp2), read(tmp3); 119 modify(tmp2, tmp2 + tmp3 - 1, 1, 1, 1, n); 120 } 121 } 122 return 0; 123 }