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
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[20]; 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=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; } 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(""); } ll n,m,q,l; ll aa[maxn]; ll ans=0; ll d[maxn]; ll C(ll l,ll r) { if(l>aa[n]) return 0; if(r<aa[0]) return 0; if(l>r) return 0; ll L,R; if(l<=0) L=1; else L=lower_bound(aa+1,aa+n+1,l)-aa; if(r>=aa[n]) R=n; else R=(upper_bound(aa+1,aa+n+1,r)-aa)-1; return R-L+1; } set<ll> S; int main() { n=read(),m=read(),q=read(),l=read(); S.insert(-inf); aa[1]=0; for(int i=2;i<=n;i++) aa[i]=read(); S.insert(inf); ll tmp=0; for(int i=1;i<=m;i++) { ll x=read(); S.insert(x); if(i==1) ans+=C(x-l,x+l); else ans+=C(max(tmp+l+1LL,x-l),x+l); tmp=x; } printf("%lld ",ans); for(int i=1;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 ",ans); S.insert(x); } }