Hotel
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Time Limit: 3000MS
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Memory Limit: 65536K
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Total Submissions: 478
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Accepted: 129
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Description
The cows are journeying north to
in 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
USACO
2008 February Gold
这是一个线段树的题目：
1. 题意建模：
关键点：找出最靠左的一个长度至少为K的线段可以用线段树来做。
对任意一个线段树中的节点(x,y)
维护3个信息：
1. (x, y) 中最长的线段的长度
2. x点开始向右的线段长度
3. y点开始向左的线段长度
题目要求我们对一个(0,
50000)长的区间做以下操作：
A. 对任意区间 (x, y) 使 F(x,
y) = 0
B. 对任意区间 (x, y) 使 F(x,
y) = 1
C. 查询任意一个区间 (x, y) 中最长的线段的长度，
D. 维护每个节点的3种信息
2.难点解析：
1.实现A, B 操作：
要把某个节点覆盖为0，或者1，不需要向下扩展。只有在需要查询到儿子区间的状态的时候才需要扩展（参考spread函数）
2.实现D的维护：
每次改变了左右儿子区间的信息的时候，就需要更新当前节点的信息(参考update函数)
3.代码实现：
#include
<stdio.h>
#define Max(a, b)
((a)>(b)?(a):(b))
const int N =
50010;
struct ST {int
i,j,m,l,r,c,lc,rc;} st[2*N]; //区间宽度的2å€
int up, n;
void bd(int d, int
x, int y) {
st[d].i = x, st[d].j = y, st[d].m =
(x+y)/2;
st[d].c = st[d].lc = st[d].rc = y-x;
if(x < y-1) {
st[d].l = ++up; bd(up, x, st[d].m);
st[d].r = ++up; bd(up, st[d].m, y);
}
}
void spread(int d)
{
if(st[d].c == st[d].j-st[d].i) {
int& l = st[d].l, &r =
st[d].r;
st[l].c = st[l].lc = st[l].rc =
st[l].j-st[l].i;
st[r].c = st[r].lc = st[r].rc =
st[r].j-st[r].i;
} else if(st[d].c == 0) {
int& l = st[d].l, &r =
st[d].r;
st[l].c = st[l].lc = st[l].rc = 0;
st[r].c = st[r].lc = st[r].rc = 0;
}
}
void update(int d)
{
st[d].c = Max(Max(st[st[d].l].c,
st[st[d].r].c), st[st[d].l].rc + st[st[d].r].lc);
if(st[st[d].l].c ==
st[st[d].l].j-st[st[d].l].i) {
st[d].lc = st[st[d].l].c +
st[st[d].r].lc;
} else st[d].lc = st[st[d].l].lc;
if(st[st[d].r].rc == st[st[d].r].j-st[st[d].r].i)
{
st[d].rc = st[st[d].r].rc +
st[st[d].l].rc;
} else st[d].rc = st[st[d].r].rc;
}
void Empty(int d,
int x, int y) {
if(x <= st[d].i && y >=
st[d].j) {
st[d].c = st[d].lc = st[d].rc = 0;
return;
}
spread(d);
if(x < st[d].m) Empty(st[d].l, x, y);
if(y > st[d].m) Empty(st[d].r, x, y);
update(d);
}
void Fill(int d,
int x, int y) {
if(x <= st[d].i && y >=
st[d].j) {
st[d].c = st[d].lc = st[d].rc =
st[d].j-st[d].i;
return;
}
spread(d);
if(x < st[d].m) Fill(st[d].l, x, y);
if(y > st[d].m) Fill(st[d].r, x, y);
update(d);
}
int Get(int d, int
l) {
spread(d);
if(st[d].c < l) return -1;
if(st[st[d].l].c >= l)
return Get(st[d].l, l);
else if(st[st[d].l].rc + st[st[d].r].lc
>= l) {
return st[st[d].l].j-st[st[d].l].rc;
}
return Get(st[d].r, l);
}
int main()
{
int i, nq, cmd, x, l, y;
up = 0;
scanf("%d%d", &n, &nq);
bd(0, 0, n);
while(nq--) {
scanf("%d", &cmd);
if(cmd == 1) {
scanf("%d", &l);
x = Get(0, l);
if(x != -1)
Empty(0, x, x+l);
printf("%d\n", x+1);
} else {
scanf("%d%d", &x,
&l);
Fill(0, x-1, x+l-1);
}
}
return 0;
}