CodeForces 46D Parking Pot（线段树区间更新）

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 pointL.
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.

Sample test(s)

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

题意：一段线性的停车场，长度为L，停车时从零点坐标开始停，要求保证新停入的车前后间距分别为f，b（在停车场的边缘不考虑车前间距或车后间距）。有n次操作，1 a表示停入一辆长度为a的车，要求输出车尾的坐标，2 a表示第a次操作中的那辆车移出停车场（注意：第a次操作不是以1开头的操作中的第a个，因为看错这个，我不停的RE......）。

明显的线段树区间更新，因为之前做过一道类似的题，保存了模板，所以这道题就直接套模板了。为了方便考虑车的前后间距，给停车场的长度加上b + f，这样就不用分类讨论了。

#include <cstdio>
#include <iostream>
#include <algorithm>
#define ls node << 1
#define rs node << 1 | 1
#define lson l, mid, ls
#define rson mid + 1, r, rs

using namespace std;

int l, b, f;
int a[105], c[105];
int sum[101205 << 2], lsum[101205 << 2], rsum[101205 << 2], visit[101205 << 2];

void pushup(int l, int r, int node)
{
int len = r - l + 1;
lsum[node] = lsum[ls];
rsum[node] = rsum[rs];
if(lsum[node] == len - (len >> 1))
lsum[node] += lsum[rs];
if(rsum[node] == (len >> 1))
rsum[node] += rsum[ls];
sum[node] = max(rsum[ls] + lsum[rs], max(sum[ls], sum[rs]));
}

void pushdown(int l, int r, int node)
{
int len = r - l + 1;
if(visit[node] != -1) {
visit[ls] = visit[rs] = visit[node];
lsum[ls] = rsum[ls] = sum[ls] = visit[node] ? 0 : len - (len >> 1);
lsum[rs] = rsum[rs] = sum[rs] = visit[node] ? 0 : (len >> 1);
visit[node] = -1;
}
}

void build(int l, int r, int node)
{
visit[node] = -1;
lsum[node] = rsum[node] = sum[node] = r - l + 1;
if(l == r)
return;
int mid = (l + r) >> 1;
build(lson);
build(rson);
}
void update(int x, int y, int v,int l, int r, int node)
{
if(x <= l&&y >= r) {
lsum[node] = rsum[node] = sum[node] = v ? 0 : r - l + 1;
visit[node] = v;
return;
}
pushdown(l, r, node);
int mid = (l + r) >> 1;
if(x <= mid)
update(x, y, v, lson);
if(y > mid)
update(x, y, v, rson);
pushup(l, r, node);
}
int query(int a, int l, int r, int node)
{
if(l == r)
return l;
pushdown(l, r, node);
int mid = (l + r) >> 1;
if(sum[ls] >= a)
return query(a, lson);
else if(rsum[ls] + lsum[rs] >= a)
return mid - rsum[ls] + 1 + b;
else return query(a, rson);
}

int main()
{
int n, choice, x, cnt = 0;
while(scanf("%d%d%d",&l, &b, &f) != EOF) {
l = l + b + f;
build(1, l, 1);
scanf("%d",&n);
while(n--) {
scanf("%d",&choice);
++cnt;
if(choice == 1) {
scanf("%d",&a[cnt]);
if(sum[1] < a[cnt] + b + f)
printf("-1\n");
else {
c[cnt] = query(a[cnt] + b + f, 1, l, 1);
update(c[cnt], c[cnt] + a[cnt] - 1, 1, 1, l, 1);
printf("%d\n", c[cnt] - 1 - b);
}
} else if(choice == 2) {
scanf("%d",&x);
update(c[x], c[x] + a[x] - 1, 0, 1, l, 1);
}
}
}
}