题目链接：115.157.200.87/front/problem?problemId=1056（HNU内网访问）

Problem Description

The ocean can be represented as the first quarter of the Cartesian plane. There are n fish in the ocean. Each fish has its own coordinates. There may be several fish at one point.

There are also m fishermen. Each fisherman has its own x-coordinate. The y-coordinate of each fisherman is equal to 0.

Each fisherman has a fishing rod of length l. Therefore, he can catch a fish at a distance less than or equal to l. The distance between a fisherman in position x and a fish in position (a, b) is |a − x| + b.

Find for each fisherman how many fish he can catch.

Input

The first line contains three integers n, m, and l (1≤n,m≤2⋅10⁵,1≤l≤10⁹) — the number of fish, the number of fishermen, and the length of the fishing rod, respectively.

Each of the next n lines contains two integersxi​ and y_i(1≤xi​,yi​,≤10⁹) — the fish coordinates.

Next line contains m integers a_i (1 ≤ a_i ≤ 10^9) — the fishermen coordinates.

Output

For each fisherman, output the number of fish that he can catch, on a separate line.

Sample Input

8 4 4
7 2
3 3
4 5
5 1
2 2
1 4
8 4
9 4
6 1 4 9

Sample Output

2
2
3
2

Note

The picture illustrates for the above example the area on which the third fisherman can catch fish.

题意：

n条鱼，m个渔夫，钓鱼线长l，渔夫的位置是(0,a_j)，和鱼(x_i,y_i)的距离是|a_i-x_i|+y_i,问每个渔夫最多能钓到多少条鱼。（要能钓到鱼需要钓鱼线长大于他们间的距离）

题解:

对于鱼(x_i,y_i)，很显然，如果y_i大于l那么那个渔夫都钓不上；否则的话，每个鱼在x轴上能被钓到的区间为[x_i-l+y_i]。转化成一个区间问题：对于每个a_j落在多少个区间内。

用left[i]表示第i个可能能被钓上的鱼在x轴上的左边界，right[i]为右边界。从小到大排序后利用二分：

找出第一个右边界大于等于a_j的序号lb，说明左边的[0,lb-1]共lb个区间不可能包含a_j;

找出第一个左边界大于a_j的序号rb，说明右边的[rb,cnt-1]共cnt-rb个区间不可能包含a_j;

对于每一个a_j，能落的区间个数即是cnt-(cnt-rb)-lb=rb-lb；（cnt为可能被钓上的鱼的个数）

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN=2e5+5;
 4 const int MAXM=2e5+5;
 5 int Left[MAXN],Right[MAXN];
 6 int a[MAXM];
 7 int read()
 8 {
 9     int s=1,x=0;
10     char ch=getchar();
11     while(!isdigit(ch)) {if(ch=='-') s=-1;ch=getchar();}
12     while(isdigit(ch)) {x=10*x+ch-'0';ch=getchar();}
13     return x*s;
14 }
15 int main()
16 {
17     int n=read(),m=read(),l=read();
18     int xi,yi,cnt=0;
19     for(int i=1;i<=n;++i)
20     {
21         xi=read(),yi=read();
22         if(yi>l) continue;
23         Left[++cnt]=xi-l+yi;
24         Right[cnt]=xi+l-yi;
25     }
26     sort(Left+1,Left+cnt+1);
27     sort(Right+1,Right+cnt+1);
28     int lb,ub,ai;
29     queue<int> ans;
30     for(int i=1;i<=m;++i)
31     {
32         ai=read();
33         lb=lower_bound(Right+1,Right+cnt+1,ai)-Right;
34         ub=upper_bound(Left+1,Left+cnt+1,ai)-Left;
35         ans.push(ub-lb);
36     }
37     while(!ans.empty())
38     {
39         cout<<ans.front()<<"\n";
40         ans.pop();
41     }
42 }

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