NEERC 2015 Southern Subregional G Hiring 树状数组

http://codeforces.com/contest/589/problem/G

G. Hiring

time limit per test

4 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

The head of human resources department decided to hire a new employee. He created a test exercise for candidates which should be accomplished in at most m working days. Each candidate has to pass this test exercise. During the j-th day a candidate is allowed to be in the office for at most tj units of time.

Overall, n candidates decided to apply for the job and sent out their resumes. Based on data received the head has defined two parameters describing every candidate: di and ri. The parameter di is the time to get prepared for work which the i-th candidate spends each morning. This time doesn't depend on day. The parameter ri is the total working time needed for the i-th candidate to accomplish the whole test exercise.

Thus the time spent in the office in the j-th day consists of di units of time to get prepared and some units of time to proceed with the exercise. A candidate can skip entire working day and do not come to the office. Obviously in this case he doesn't spend di units of time to prepare.

To complete the exercise a candidate should spend exactly ri units of time working on the exercise (time to prepare is not counted here).

Find out for each candidate what is the earliest possible day when he can fully accomplish the test exercise. It is allowed to skip working days, but if candidate works during a day then he must spend di units of time to prepare for work before he starts progressing on the exercise.

Input

The first line contains two integer numbers n,  m (1 ≤ n,  m ≤ 2·105) — the number of candidates and the maximum number of working days to do the test exercise.

The second line contains m integer numbers t1, t2, ..., tm (1 ≤ tj ≤ 106) — the durations of working days in time units.

The following n lines contain two integers each: di,  ri (0 ≤ di ≤ 106,  1 ≤ ri ≤ 106) — how much time in the beginning of a day is required for i-th candidate before he starts his work on the test exercise and how much time it is needed for him to accomplish this task.

Output

Output a sequence of n integer numbers b1, b2, ..., bn, where bi is the earliest day when the i-th candidate can finish the test exercise.

In case the i-th candidate cannot finish the test exercise in m days output bi = 0.

Days in this problem are numbered from 1 to m in the order they are given in the input.

Sample test(s)

input

3 3
4 2 5
1 3
2 5
3 4

output

1 3 0 

/**
NEERC 2015 Southern Subregional G Hiring   树状数组
题目大意：有n个应聘者，每个人必须做完a[i]的工作量，但是每天他们投入工作前必须热身b[i]时间，每天每个人的工作量
           最大为t[i],问每个人最早哪天完工。
解题思路：很好的题目，二分第几天mid,用一个树状数组求到mid天的除去准备时间的总时间，另一个树状数组求1到mid天有几天因为一天
          的时间比准备时间少而删去的天数

*/
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include <iostream>
#include <vector>
using namespace std;
typedef long long LL;
const int maxn=1e6+7;

LL sum[2][maxn];
int n,m,num[maxn];
vector<int>vec[maxn];

struct note
{
    int x,y,id;
    bool operator <(const note &other)const
    {
        return x<other.x;
    }
} a[maxn];

int lowbit(int x)
{
    return x & -x;
}

LL query(int t,int x)
{
    LL ret=0;
    while(x>0)
    {
        ret+=sum[t][x];
        x-=lowbit(x);
    }
    return ret;
}

void add(int t,int x,int d=1)
{
    while(x<=m)
    {
        sum[t][x]+=d;
        x+=lowbit(x);
    }
}

int get(int t)
{
    int l=1,r=m,ans=0;
    while(l<=r)
    {
        int mid=(l+r)/2;
        LL tmp=mid-query(1,mid);
        LL cnt=query(0,mid)-tmp*a[t].x;
        if(cnt>=a[t].y)
        {
            ans=mid;
            r=mid-1;
        }
        else
            l=mid+1;
    }
    return ans;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(sum,0,sizeof(sum));
        for(int i=0; i<maxn; i++)vec[i].clear();
        for(int i=1; i<=m; i++)
        {
            int x;
            scanf("%d",&x);
            add(0,i,x);
            vec[x].push_back(i);
        }
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&a[i].x,&a[i].y);
            a[i].id=i;
        }
        sort(a,a+n);
        int pos=0;
        for(int i=0; i<n; i++)
        {
            while(pos<=a[i].x)
            {
                for(int j=0; j<vec[pos].size(); j++)
                {
                    int v=vec[pos][j];
                    add(0,v,-pos);
                    add(1,v);
                }
                pos++;
            }
            num[a[i].id]=get(i);
        }
        for(int i=0; i<n; i++)printf(i==n-1?"%d\n":"%d ",num[i]);
    }
    return 0;
}