【CROC 2016 — QualificationB】【队列模拟】Processing Queries 按时间顺序处理任务 队列长度不超过siz

B. Processing Queries

time limit per test
5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

In this problem you have to simulate the workflow of one-thread server. There are n queries
to process, the i-th will be received at moment ti and
needs to be processed for di units
of time. All ti are
guaranteed to be distinct.
When a query appears server may react in three possible ways:

If server is free and query queue is empty, then server immediately starts to process this query.

If server is busy and there are less than b queries in the queue, then new query
is added to the end of the queue.

If server is busy and there are already b queries pending in the queue, then new
query is just rejected and will never be processed.
As soon as server finished to process some query, it picks new one from the queue (if it's not empty, of course). If a new query comes at some moment x,
and the server finishes to process another query at exactly the same moment, we consider that first query is picked from the queue and only then new query appears.
For each query find the moment when the server will finish to process it or print -1 if
this query will be rejected.

Input
The first line of the input contains two integers n and b (1 ≤ n, b ≤ 200 000) —
the number of queries and the maximum possible size of the query queue.
Then follow n lines
with queries descriptions (in chronological order). Each description consists of two integers ti and di(1 ≤ ti, di ≤ 109),
where ti is
the moment of time when the i-th query appears and di is
the time server needs to process it. It is guaranteed that ti - 1 < ti for
all i > 1.

Output
Print the sequence of n integers e1, e2, ..., en,
where ei is
the moment the server will finish to process the i-th query (queries are numbered in the order
they appear in the input) or  - 1 if the corresponding query will be rejected.

Examples

input

5 1
2 9
4 8
10 9
15 2
19 1

output

11 19 -1 21 22

input

4 1
2 8
4 8
10 9
15 2

output

10 18 27 -1

Note
Consider the first sample.

The server will start to process first query at the moment 2 and will finish to process it at the moment 11.

At the moment 4 second query appears and proceeds to the queue.

At the moment 10 third query appears. However, the server is still busy with query 1, b = 1 and
there is already query 2 pending in the queue, so third query is just rejected.

At the moment 11 server will finish to process first query and will take the second query from the queue.

At the moment 15 fourth query appears. As the server is currently busy it proceeds to the queue.

At the moment 19 two events occur simultaneously: server finishes to proceed the second query and the fifth query
appears. As was said in the statement above, first server will finish to process the second query, then it will pick the fourth query from the queue and only then will the fifth query appear. As the queue is empty fifth query is proceed there.

Server finishes to process query number 4 at the moment 21.
Query number 5 is picked from the queue.

Server finishes to process query number 5 at the moment 22.

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 2e5+10, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int casenum, casei;
int n, siz;
struct A
{
LL t, d;
}a
;
queue<int>q;
LL ans
;
int main()
{
while(~scanf("%d%d",&n,&siz))
{
for (int i = 1; i <= n; ++i)
{
scanf("%lld%lld", &a[i].t,&a[i].d);
}
LL T = 0;//T表示上个任务的完成时间
while (!q.empty())q.pop();
for (int i = 1; i <= n; ++i)
{
//我们把a[i].t之前时间的任务先做完
while(!q.empty())
{
int o = q.front();
if (T <= a[i].t)
{
T = ans[o] = max(T, a[o].t) + a[o].d;
q.pop();
}
else break;
}
//如果队列空了，我们直接处理这个任务
if (T<=a[i].t&&q.empty())
{
T = ans[i] = a[i].t + a[i].d;
continue;
}
//如果队列没空，我们考虑加入队列
if (q.size() < siz)q.push(i);
else ans[i] = -1;
}
while (!q.empty())
{
int o = q.front();
LL finish = max(T, a[o].t) + a[o].d;
T = ans[o] = finish;
q.pop();
}
for (int i = 1; i <= n; ++i)printf("%lld ", ans[i]);
puts("");
}
return 0;
}
/*
【trick&&吐槽】
小心爆int

【题意】
有n（2e5）个询问。队列上限为b（2e5）
询问按照时间顺序给你。
每个询问的时间都不同，分别是ti
每个询问需要处理的时间分别是di
对于每个询问。
1，如果机器空闲，且没有要处理的询问，则立即处理。
2，如果机器工作，且要处理的询问数小于b，则加入队尾
3，如果机器工作，且要处理的询问数等于b，则丢弃任务。
当机器处理完某个询问，它会从队列中选择一个新的做处理。
如果这时刚好有新任务，这个任务的优先度在后。
输出每个任务的完成时间。

【类型】
队列模拟

【分析】
我们只要模拟就好了。
在处理每个工作到来的时候，先把对列中在它时间之间的处理。
这样才能决定这个工作的目标状态。

*/