【26.8%】【CF 46D】Parking Lot

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Nowadays it is becoming increasingly difficult to park a car in cities successfully. Let's imagine a segment of a street as long as L meters along which a parking lot is located. Drivers should park their cars strictly parallel to the pavement on the right side of the street (remember that in the country the authors of the tasks come from the driving is right side!). Every driver when parking wants to leave for themselves some extra space to move their car freely, that's why a driver is looking for a place where the distance between his car and the one behind his will be no less than b meters and the distance between his car and the one in front of his will be no less than f meters (if there's no car behind then the car can be parked at the parking lot segment edge; the same is true for the case when there're no cars parked in front of the car). Let's introduce an axis of coordinates along the pavement. Let the parking lot begin at point 0 and end at point L. The drivers drive in the direction of the coordinates' increasing and look for the earliest place (with the smallest possible coordinate) where they can park the car. In case there's no such place, the driver drives on searching for his perfect peaceful haven. Sometimes some cars leave the street and free some space for parking. Considering that there never are two moving cars on a street at a time write a program that can use the data on the drivers, entering the street hoping to park there and the drivers leaving it, to model the process and determine a parking lot space for each car.

Input

The first line contains three integers L, b и f (10 ≤ L ≤ 100000, 1 ≤ b, f ≤ 100). The second line contains an integer n (1 ≤ n ≤ 100) that indicates the number of requests the program has got. Every request is described on a single line and is given by two numbers. The first number represents the request type. If the request type is equal to 1, then in that case the second number indicates the length of a car (in meters) that enters the street looking for a place to park. And if the request type is equal to 2, then the second number identifies the number of such a request (starting with 1) that the car whose arrival to the parking lot was described by a request with this number, leaves the parking lot. It is guaranteed that that car was parked at the moment the request of the 2 type was made. The lengths of cars are integers from 1 to 1000.

Output

For every request of the 1 type print number -1 on the single line if the corresponding car couldn't find place to park along the street. Otherwise, print a single number equal to the distance between the back of the car in its parked position and the beginning of the parking lot zone.

Examples

input

30 1 2
6
1 5
1 4
1 5
2 2
1 5
1 4

output

0
6
11
17
23

input

30 1 1
6
1 5
1 4
1 5
2 2
1 5
1 4

output

0
6
11
17
6

input

10 1 1
1
1 12

output

-1

【题解】

把车开走操作的x指的是n个操作里面的第x个操作对应的车。

这题好坑啊，也怪自己没看清题目。它说的是长度为L。。

0..L说的是点。然后相邻两个点之间的距离为1，这样就组成长度为L的一段路了-_-.

我们不要管他。就变成0..L-1

这样每个点就代表一个距离了。而不是真的是一个点。。

然后我们在这条路的前边扩充b，在后边扩充f

整个区间就变成[-b..L-1+f]了(线段树不管你区间是不是负都可以的^_^)

输入的车的长度是x

那么问题就转换成在这个区间内找一段x+b+f的连续空位置。

假设找到的最左端的位置是pos;

那么pos+b就是要输出的答案了。

然后占据的时候不是占据pos..pos+b+f+x

而应该占据pos+b..pos+x-1

但我觉得这个问题还是有点BUG的。

就是后面进来的车停在了前面进来的车的前面b单位长度处。那么如果f>b，前面那辆车不就不满足要求了吗。。。(路人:出题人就是爷，你管他呢)

找连续空位置的话。

记录llx[rt],rlx[rt],lx[rt]分别表示这个节点从最左开始连续的空位置数目，这个节点从最右开始连续的空位置数目，整个区间不管哪里，连续的空位置数目。

然后所求的连续空位置有3种可能。

1.全部在左区间

2.全部在右区间。

3.横跨两个区间。

所以得到lx[rt] = max(lx[rt<<1],lx[rt<<1|1],rlx[rt<<1]+llx[rt<<1|1]);

因为连续区间块的时候要尽量往左。

所以先递归左儿子。然后是横跨中间的情况(如果是这种情况就可以直接输出起始位置了)；最后是右儿子；

【代码】

#include <cstdio>
#include <algorithm>
#define lson begin,m,rt<<1
#define rson m+1,end,rt<<1|1

using namespace std;

const int MAXL = 101000;

struct data2
{
	int l,r;
};

data2 qujian[101];
int l, b, f, n;
int cover[MAXL * 4],llx[MAXL*4],rlx[MAXL*4],lx[MAXL*4];

void push_up(int rt,int len)
{
	lx[rt] = max(lx[rt << 1], lx[rt << 1 | 1]);
	lx[rt] = max(lx[rt], rlx[rt << 1] + llx[rt << 1 | 1]);
	llx[rt] = llx[rt << 1];
	if (llx[rt] == (len - (len >> 1))) //如果左区间全是连续的空位置。
		llx[rt] += llx[rt << 1 | 1];//则加上右区间从最左开始的连续空位置数目。
	rlx[rt] = rlx[rt << 1 | 1];
	if (rlx[rt] == (len >> 1))
		rlx[rt] += rlx[rt << 1];
}

void build(int begin, int end, int rt)
{
	cover[rt] = 0;
	if (begin == end)
	{
		llx[rt] = rlx[rt] = lx[rt] = 1;
		return;
	}
	int m = (begin + end) >> 1;
	build(lson);
	build(rson);
	push_up(rt,end-begin+1);
}

void input_data()
{
	scanf("%d%d%d", &l, &b, &f);
	build(-b, l + f -1, 1);
}

void push_down(int rt,int len)
{
	if (cover[rt] != -1)
	{
		cover[rt << 1] = cover[rt << 1 | 1] = cover[rt];
		if (cover[rt] == 1)
		{
			llx[rt << 1] = rlx[rt << 1] = lx[rt << 1] = 0;
			llx[rt << 1| 1] = rlx[rt << 1 | 1] = lx[rt << 1 | 1] = 0;
		}
		else
		{
			llx[rt << 1] = rlx[rt << 1] = lx[rt << 1] = len - (len >> 1);
			llx[rt << 1 | 1] = rlx[rt << 1 | 1] = lx[rt << 1 | 1] = len >> 1;
		}
		cover[rt] = -1;
	}
}

int query(int len, int begin, int end, int rt)
{
	if (begin == end)
		return begin;
	push_down(rt,end - begin+1);
	int m = (begin + end) >> 1;
	if (lx[rt << 1] >= len)
		return query(len, lson);
	else
		if (rlx[rt << 1] + llx[rt << 1 | 1] >= len)
			return m - rlx[rt << 1] + 1; //返回的这个坐标可以手算模拟下。
		else
			return query(len, rson);
}

void up_data(int l, int r, int num, int begin, int end, int rt) //用于更新节点。
{
	if (l <= begin && end <= r)
	{
		cover[rt] = num;
		if (num == 1)
		{
			llx[rt] = rlx[rt] = lx[rt] = 0;
			return;
		}
		else
		{
			llx[rt] = rlx[rt] = lx[rt] = end - begin + 1;
		}
		return;
	}
	push_down(rt,end - begin+1);
	int m = (begin + end) >> 1;
	if (l <= m)
		up_data(l, r, num, lson);
	if (m < r)
		up_data(l, r, num, rson);
	push_up(rt, end - begin + 1);
}

void output_ans()
{
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		int op, x;
		scanf("%d%d", &op, &x);
		if (op == 1)
		{
			int len = x + b + f;
			if (len > lx[1])
				printf("-1\n");
			else
			{
				int qidian = query(len, -b, l + f - 1, 1);
				int cl, cr;
				cl = qidian + b; //[cl,cr]是要修改的区间。
				cr = cl + x - 1;
				qujian[i].l = cl;
				qujian[i].r = cr;
				up_data(cl, cr, 1, -b, l + f - 1, 1);
				printf("%d\n", qidian+b);
			}
		}
		else
		{
			int zuo, you;
			zuo = qujian[x].l;
			you = qujian[x].r;
			up_data(zuo, you, 0, -b, l + f - 1, 1);
		}
	}
}

int main()
{
	//freopen("F:\\rush.txt", "r", stdin);
	//freopen("F:\\rush_out.txt", "w", stdout);
	input_data();
	output_ans();
	return 0;
}