Time Limit:3000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
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 Dicontiguous 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
/****
思路： 线段树的区间合并问题： 要查询连续区间长度，所以要记录最长的连续区间，一段区间的连续可以分为左连续，右连续，中间连续，然后记录就可以了 父节点左连续区间为左儿子左连续，如果左儿子在整个左区间内连续则再加上右儿子左连续 父节点右连续区间为右儿子右连续，如果右儿子在整个右区间内连续则再加上左儿子右连续 父节点中间连续区间为左儿子右连续加上右儿子左连续 然后记录每个节点总的最长连续区间的值 每次向上更新或者向下延迟都要重新计算节点信息
****/
#include"iostream"
#include"algorithm"
#include"cstdio"
#include"cstring"
#include"cmath"
//#define max(a,b) a>b?a:b //【这个地方WA了好多次。。。】
//#define min(a,b) a<b?a:b //【这个宏定义有问题。。。以后再也不随便用宏定义了。。。WA哭了。。。】
#define MX 110000
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std; int lsum[MX<<],rsum[MX<<],sum[MX<<];
int lazy[MX<<]; // 1 a 询问是否存在a间连续的空房间，有则住进最左边 【】
// 2 a b 将 [a,a+b-1] 编号的房间清空 【成段更新】 void PushDown(int rt,int m) {
if(lazy[rt]!=INF) {//1 ->
lazy[rt<<]=lazy[rt<<|]=lazy[rt]; //下移懒惰标记
sum[rt<<]=lsum[rt<<]=rsum[rt<<]=lazy[rt]*(m-(m>>)); //因为
sum[rt<<|]=lsum[rt<<|]=rsum[rt<<|]=lazy[rt]*(m>>);
lazy[rt]=INF;
}
} void PushUp(int rt,int m) {
lsum[rt]=lsum[rt<<];
rsum[rt]=rsum[rt<<|];
if(lsum[rt]==m-(m>>)) lsum[rt]+=lsum[rt<<|]; // 这里lson[rt<<1]是表示的左区间的左最大值,lson[rt<<1|1]表示的是左区间的右最大值
if(rsum[rt]== (m>>) ) rsum[rt]+=rsum[rt<<]; // rson[。。。] 类似 而左区间的右最大值加右区间的左最大值就是中间区间的最大值
sum[rt]=max(max(sum[rt<<],sum[rt<<|]),lsum[rt<<|]+rsum[rt<<]); //保存两边子节点和 中最大值 中的最大值
} void Build(int l,int r,int rt) {
sum[rt]=lsum[rt]=rsum[rt]=r-l+; //把每一个空房间标记为1，每个父节点记录为子节点最大的空区间
lazy[rt]=INF; //懒惰标记清空
if(l==r) return ;
int m=(r+l)>>;
Build(lson);
Build(rson);
} void UpData(int L,int R,int val,int l,int r,int rt) {
if(L<=l&&r<=R) {
sum[rt]=lsum[rt]=rsum[rt]= val ? (r-l+):;
lazy[rt]=val;
return ;
}
PushDown(rt,r-l+);
int m=(r+l)>>;
if(L<=m) UpData(L,R,val,lson);
if(R> m) UpData(L,R,val,rson);
PushUp(rt,r-l+);
} int Query(int ll,int l,int r,int rt) {
if(l==r) return l;
PushDown(rt,r-l+);
int m=(r+l)>>;
if(sum[rt<<]>=ll) return Query(ll,lson); //如果左最大值满足就在往左继续搜
else if(rsum[rt<<]+lsum[rt<<|]>=ll)return m-rsum[rt<<]+;//如果中最大值，满足就输出左区间的右最大值的第一个位置
return Query(ll,rson); //否则就向右继续搜
} int main() {
int n,m;
scanf("%d%d",&n,&m);
Build(,n,);
int o,a,b;
for(int i=; i<m; i++) {
scanf("%d",&o) ;
if(o==) {
scanf("%d",&a);
if(sum[]<a) printf("0\n");
else {
int q=Query(a,,n,);
printf("%d\n",q);
UpData(q,q+a-,,,n,);
}
} else if(o==) {
scanf("%d%d",&a,&b);
UpData(a,a+b-,,,n,);
}
}
return ;
}