Codeforces Gym 100187M M. Heaviside Function two pointer

M. Heaviside Function

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100187/problem/M

Description

Heaviside function is defined as the piecewise constant function whose value is zero for negative argument and one for non-negative argument:

You are given the function f(x) = θ(s1x - a1) + θ(s2x - a2) + ... + θ(snx - an), where si =  ± 1. Calculate its values for argument values x1, x2, ..., xm.

Input

The first line contains a single integer n (1 ≤ n ≤ 200000) — the number of the summands in the function.

Each of the next n lines contains two integers separated by space — si and ai (si =  ± 1,  - 109 ≤ ai ≤ 109) — parameters of the i-th summand.

The next line contains a single integer m (1 ≤ m ≤ 200000) — the number of the argument values you should calculate the value of the function for.

The last line contains m integers x1, ..., xm ( - 109 ≤ xi ≤ 109) separated by spaces — the argument values themselves.

Output

Output m lines. i-th line should contain the value of f(xi).

Sample Input

6
1 3
-1 2
1 9
-1 2
1 7
-1 2
8
0 12 2 8 4 -3 7 9

Sample Output

0
3
0
2
1
3
2
3

HINT

 

题意

if(sx-a>=0)ans++;问你每个数通过这个公式，最后的ans是多少

题解：

当s=1时，很显然这个直线是单调向上的，我们可以利用two pointer来优化就好了

复杂度O(n+m)

代码

#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)
#define maxn 200101
#define mod 1000000009
#define eps 1e-9
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//**************************************************************************************

struct node
{
    int x,y;
};

bool cmp1(node a,node b)
{
    return a.x<b.x;
}
bool cmp2(node a,node b)
{
    return a.y<b.y;
}
node a[maxn];
vector<int> a1;
vector<int> a2;
node b[maxn];
int ans[maxn];

int main()
{
    int flag1=0,flag2=0;
    int n=read();
    for(int i=0;i<n;i++)
    {
        a[i].x=read(),a[i].y=read();
        if(a[i].x==1)
            flag1++,a1.push_back(a[i].y);
        else
            flag2++,a2.push_back(a[i].y);
    }
    int m=read();
    for(int i=0;i<m;i++)
        b[i].x=read(),b[i].y=i;
    sort(b,b+m,cmp1);
    sort(a1.begin(),a1.end());
    sort(a2.begin(),a2.end());

    for(int i=m-1;i>=0;i--)
    {
        if(flag1==0)
            break;
        while(b[i].x<a1[flag1-1]&&flag1>0)
            flag1--;
        if(flag1==0)
            break;
        ans[b[i].y]+=flag1;
    }
    for(int i=0;i<m;i++)
    {
        if(flag2==0)
            break;
        while(b[i].x>-a2[flag2-1]&&flag2>0)
            flag2--;
        if(flag2==0)
            break;
        ans[b[i].y]+=flag2;
    }
    for(int i=0;i<m;i++)
        printf("%d\n",ans[i]);
}