Codeforces Gym 100523E E - Gophers SET

E - Gophers
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87794#problem/E

Description

Dick Dastardly wants to bedevil poor Bytean gophers. These nice little creatures live in holes in the upper parts of High Bytemountains. Dick has found a mountain ridge with n gopher holes located along a straight line (for simplicity, we index the holes from 1 to n, from west to east). Dick plans to torture gophers using rock & roll music. He has bought m CD players, put a different Bytels’ album in each of them and arranged the CD players along the ridge. The music from a CD player disturbs gophers located in holes distant by at most l meters from it. Feeling troubled, the gophers asked you to check in which holes they will not be able to sleep well during this winter. But now Dick Dastardly wants to make even more mess. . . He will move the CD players from time to time. The gophers were able to steal Dick’s secret plan and now they know precisely that on the morning of the i-th day Dick will take the CD player located pi meters from the hole number 1 and will put it at a point located ri meters from that hole. Help the gophers and count the number of holes in which they will not be able to fall asleep after each such operation.

Input

The first line of input contains four integers n, m, d and l (2 ¬ n; m ¬ 500 000, 1 ¬ d ¬ 500 000, 1 ¬ l ¬ 109 ) representing the number of gophers’ holes, the number of Dick’s CD players, the number of days and the range of a CD player, respectively. The second line of input contains n − 1 integers x2; x3; : : : ; xn (0 < x2 < x3 < : : : < xn ¬ 109 ) denoting the distances of the holes 2; 3; : : : ; n from the hole number 1. The third line contains m integers z1; z2; : : : ; zm (0 ¬ z1 < z2 < : : : < zm ¬ 109 ) denoting the distances of the consecutive CD players from the hole number 1. All the CD players are located to the east of this hole. Next, d lines follow. The i-th of these lines contains two integers pi and ri (0 ¬ pi ; ri ¬ 109 , pi 6= ri ) meaning that in the beginning of the i-th day Dick is going to move the CD player located pi meters from the hole number 1 to the point located ri meters to the east from that hole. You may assume that before every such operation there is a CD player at the position pi and there are no CD players at the position ri .

Output

Your program should output d + 1 lines. The line number i (for i = 1; 2; : : : ; d) should contain one integer representing the number of holes in which no gopher would be able to sleep well during the night before the i-th Dick’s operation. The last line should contain this number after the last Dick’s operation.

Sample Input

5 3 4 1 2 5 6 11 2 4 8 2 1 4 10 8 6 1 8

Sample Output

2 3 3 5 3

HINT

题意

给你n个点，m个长度为l的线段

有Q次询问，每次询问就是把x位置的线段挪到y位置，然后问你这些线段覆盖了多少个点

题解：

首先，我们知道每一个线段的长度都是一样的，而且题目给了，没有任何两条线段是在同一个点的，于是我们就可以用set做

对于每一个线段，他覆盖的区域实际上是[max(a[i-1]+l,a[i]-l),min(a[i+1]-l,a[i]+l)]这个区域

然后我们用set去维护就好了

代码：

//qscqesze
#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 maxn 1000001
#define mod 10007
#define eps 1e-9
int Num;
char CH[];
const int inf=0x7fffffff;   //нчоч╢С
//const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
//**************************************************************************************
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=;if(!x){putchar('');puts("");return;}
    while(x>)CH[++Num]=x%,x/=;
    while(Num)putchar(CH[Num--]+);
    puts("");
}
ll n,m,q,l;
ll aa[maxn];
ll ans=;
ll d[maxn];
ll C(ll l,ll r)
{
    if(l>aa[n])
        return ;
    if(r<aa[])
        return ;
    if(l>r)
        return ;
    ll L,R;
    if(l<=)
        L=;
    else
        L=lower_bound(aa+,aa+n+,l)-aa;
    if(r>=aa[n])
        R=n;
    else
        R=(upper_bound(aa+,aa+n+,r)-aa)-;
    return R-L+;
}
set<ll> S;
int main()
{
    n=read(),m=read(),q=read(),l=read();
    S.insert(-inf);
    aa[]=;
    for(int i=;i<=n;i++)
        aa[i]=read();
    S.insert(inf);
    ll tmp=;
    for(int i=;i<=m;i++)
    {
        ll x=read();
        S.insert(x);
        if(i==)
            ans+=C(x-l,x+l);
        else
            ans+=C(max(tmp+l+1LL,x-l),x+l);
        tmp=x;
    }
    printf("%lld\n",ans);
    for(int i=;i<=q;i++)
    {
        ll x=read();
        ll c=*++S.lower_bound(x);
        ll d=*--S.lower_bound(x);
        ans-=C(max(d+l+1LL,x-l),min(c-l-1LL,x+l));
        int e=*S.lower_bound(x);
        S.erase(e);
        x=read();
        c=*S.lower_bound(x);
        d=*--S.lower_bound(x);
        ans+=C(max(d+l+1LL,x-l),min(c-l-1LL,x+l));
        printf("%lld\n",ans);
        S.insert(x);
    }
}