Lyft Level 5 Challenge 2018 - Final Round (Open Div. 2) B 1075B （思维）

B. Taxi drivers and Lyft

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Palo Alto is an unusual city because it is an endless coordinate line. It is also known for the office of Lyft Level 5.

Lyft has become so popular so that it is now used by all mm taxi drivers in the city, who every day transport the rest of the city residents — nn riders.

Each resident (including taxi drivers) of Palo-Alto lives in its unique location (there is no such pair of residents that their coordinates are the same).

The Lyft system is very clever: when a rider calls a taxi, his call does not go to all taxi drivers, but only to the one that is the closest to that person. If there are multiple ones with the same distance, then to taxi driver with a smaller coordinate is selected.

But one morning the taxi drivers wondered: how many riders are there that would call the given taxi driver if they were the first to order a taxi on that day? In other words, you need to find for each taxi driver ii the number aiai  — the number of riders that would call the ii -th taxi driver when all drivers and riders are at their home?

The taxi driver can neither transport himself nor other taxi drivers.

Input

The first line contains two integers nn and mm (1≤n,m≤1051≤n,m≤105 ) — number of riders and taxi drivers.

The second line contains n+mn+m integers x1,x2,…,xn+mx1,x2,…,xn+m (1≤x1<x2<…<xn+m≤1091≤x1<x2<…<xn+m≤109 ), where xixi is the coordinate where the ii -th resident lives.

The third line contains n+mn+m integers t1,t2,…,tn+mt1,t2,…,tn+m (0≤ti≤10≤ti≤1 ). If ti=1ti=1 , then the ii -th resident is a taxi driver, otherwise ti=0ti=0 .

It is guaranteed that the number of ii such that ti=1ti=1 is equal to mm .

Output

Print mm integers a1,a2,…,ama1,a2,…,am , where aiai is the answer for the ii -th taxi driver. The taxi driver has the number ii if among all the taxi drivers he lives in the ii -th smallest coordinate (see examples for better understanding).

Examples

Input

Copy

3 1
1 2 3 10
0 0 1 0

Output

Copy

3 

Input

Copy

3 2
2 3 4 5 6
1 0 0 0 1

Output

Copy

2 1 

Input

Copy

1 4
2 4 6 10 15
1 1 1 1 0

Output

Copy

0 0 0 1 


看到这道题感觉很简单就是写不出来，队友都可快做出来了，后来看了大佬代码，真是美如画。

正反遍历预处理每个点的最近司机，然后再遍历判断即可

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
using namespace std;
#define debug(x) cout << "&&" << x << "&&" << endl;
#define lowbit(x) (x&-x)
#define mem(a,b) memset(a, b, sizeof(a));
typedef long long ll;
const ll mod=1000000007;
const int inf = 0x3f3f3f3f;
void read() {freopen("in.txt","r",stdin);}
ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
//head

const int maxn=200010;
int n,m,L[maxn],R[maxn],ans[maxn],pos[maxn],dir[maxn];
int main() {
    //read();
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n+m;i++) scanf("%d",&pos[i]);
    for(int i=1;i<=n+m;i++) scanf("%d",&dir[i]);
    for(int i=1;i<=n+m;i++) {//从1开始，很巧妙，预处理也很巧
        if(dir[i]) L[i]=i;
        else L[i]=L[i-1];
    }
    R[n+m+1]=n+m+1;
    for(int i=n+m;i>=1;i--) {
        if(dir[i]) R[i]=i;
        else R[i]=R[i+1];
    }
    for(int i=1;i<=n+m;i++) {
        if(!dir[i]) {//如果是乘客，并且L[i]为0即没有左司机，或右司机>n+m并且左距离大于右距离
            if(!L[i]||(R[i]<=n+m&&pos[i]-pos[L[i]]>pos[R[i]]-pos[i])) ans[R[i]]++;
            else ans[L[i]]++;
        }
    }
    for(int i=1;i<=n+m;i++) {
        if(dir[i]) printf("%d ",ans[i]);
    }
}