http://poj.org/problem?id=3667
| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 9484 | Accepted: 4066 |
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
{
int l,r;
int ml,mm,mr; //左边的空位，区间最大的空位，右边的空位
}node[N<<2];
1 /** 2 judge status: Accepted exe.time:563MS 3 exe.memory 2712K language:C++ 4 */ 5 6 #include<iostream> 7 #include<stdio.h> 8 #include<string.h> 9 #include<algorithm> 10 11 #define N 50010 12 13 #define lson p<<1 14 #define rson p<<1|1 15 16 using namespace std; 17 18 struct Nod 19 { 20 int l,r; 21 int ml,mm,mr; //左边的空位，区间最大的空位，右边的空位 22 }node[N<<2]; 23 24 int id; 25 26 void building(int l,int r,int p)//建树 27 { 28 node[p].l = l; 29 node[p].r = r; 30 node[p].mm=node[p].mr=node[p].ml= r-l+1; 31 if(l==r) return; 32 int mid = (l+r)>>1; 33 building(l,mid,lson); 34 building(mid+1,r,rson); 35 } 36 37 void judgeChange(int p) //向下更新，状态延迟标记，当搜索到整个区间一样时，向下更新一层，然后进行操作 38 { 39 if(node[p].mm==node[p].r-node[p].l+1) 40 { 41 node[lson].ml=node[lson].mm=node[lson].mr=node[lson].r-node[lson].l+1; 42 node[rson].ml=node[rson].mm=node[rson].mr=node[rson].r-node[rson].l+1; 43 } 44 if(node[p].mm==0) 45 { 46 node[lson].ml=node[lson].mm=node[lson].mr=0; 47 node[rson].ml=node[rson].mm=node[rson].mr=0; 48 } 49 } 50 51 void findId(int p,int val) //查找可以开房间的起始位置 52 { 53 if(node[p].l==node[p].r&&val==1) //当val为1并且已经达到叶子节点，说明没有孩子了，不必再往下查找 54 { 55 id = node[p].l; 56 return; 57 } 58 judgeChange(p); //状态延迟标记更新 59 if(node[p].mm>=val) //区间最大连续>=val时 60 { 61 if(node[lson].mm>=val) //优先左边搜索 62 { 63 findId(lson,val); 64 } 65 else if(node[rson].ml+node[lson].mr>=val) //然后中间 66 { 67 id = node[lson].r - node[lson].mr + 1; 68 } 69 else if(node[rson].mm>=val) //最后右边 70 { 71 findId(rson,val); 72 } 73 } 74 } 75 76 void Up(int p) //向上更新状态 77 { 78 node[p].ml = node[lson].ml; 79 node[p].mr = node[rson].mr; 80 if(node[lson].ml==node[lson].r-node[lson].l+1) 81 { 82 node[p].ml+=node[rson].ml; 83 } 84 if(node[rson].mr==node[rson].r-node[rson].l+1) 85 { 86 node[p].mr+=node[lson].mr; 87 } 88 node[p].mm = max(node[lson].mm,node[rson].mm); 89 node[p].mm = max(node[p].mm,node[lson].mr+node[rson].ml); 90 node[p].mm = max(node[p].mm,node[p].ml); 91 node[p].mm = max(node[p].mm,node[p].mr); 92 } 93 94 void update(int l,int r,int p,int op) //更新 95 { 96 if(node[p].l==l&&node[p].r==r) 97 { 98 if(op==1) node[p].ml=node[p].mm=node[p].mr=0; //执行操作1 99 else node[p].ml=node[p].mm=node[p].mr=r-l+1; //执行操作2 100 return; 101 } 102 judgeChange(p); //状态延迟标记更新 103 int mid = (node[p].l+node[p].r)>>1; 104 if(r<=mid) update(l,r,lson,op); 105 else if(l>mid) update(l,r,rson,op); 106 else 107 { 108 update(l,mid,lson,op); 109 update(mid+1,r,rson,op); 110 } 111 Up(p); //由左右孩子节点向上更新ml，mm，mr值 112 } 113 114 115 int main() 116 { 117 int n,m; 118 scanf("%d%d",&n,&m); 119 building(1,n,1); 120 while(m--) 121 { 122 int op,a,b; 123 scanf("%d",&op); 124 if(op==1) 125 { 126 scanf("%d",&a); 127 id = 0; 128 findId(1,a); //查找可插入位置 129 printf("%d ",id); 130 if(id) update(id,id+a-1,1,1); //在id位置插入a个数，即在[id,id+a-1]插入 131 } 132 else if(op==2) 133 { 134 scanf("%d%d",&a,&b); 135 update(a,a+b-1,1,2); //在a位置删除b个数，即把在[a,a+b-1]区间清空 136 } 137 } 138 return 0; 139 }