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.
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.
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.
30 1 2 6 1 5 1 4 1 5 2 2 1 5 1 4
0 6 11 17 23
30 1 1 6 1 5 1 4 1 5 2 2 1 5 1 4
0 6 11 17 6
10 1 1 1 1 12
-1
題意：一段線性的停車場，長度爲L，停車時從零點座標開始停，要求保證新停入的車前後間距分別爲f，b（在停車場的邊緣不考慮車前間距或車後間距）。有n次操作，1 a表示停入一輛長度爲a的車，要求輸出車尾的座標，2 a表示第a次操作中的那輛車移出停車場（注意：第a次操作不是以1開頭的操作中的第a個，因爲看錯這個，我不停的RE......）。
明顯的線段樹區間更新，因爲之前做過一道類似的題，保存了模板，所以這道題就直接套模板了。爲了方便考慮車的前後間距，給停車場的長度加上b + f，這樣就不用分類討論了。
#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);
}
}
}
}
