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Spoj-BGSHOOT

2024-10-26 14:59:39   209人阅读

The problem is about Mr.BG who is a great hunter. Today he has gone to a dense forest for hunting and killing animals.

Sadly, he has only one bullet in his gun. He wants to kill as many animals as possible with only one bullet.

He has already known the information of duration availability of all animals of the forest.

So, he is planning to shoot at a time so that he could kill maximum animal. 

Input

Input begins with an integer N denoting total numbers of animals.

Next N lines contains the duration of availability of animal denoting by X (Starting time) and Y (Ending time) .

Then, there will be Q, denoting the total numbers of queries to be answer.

Each query giving two integer L and R, L denoting the time hunter will come to forest and begins shooting

and R denoting last time upto which he will stay at forest for hunting.

Output

For each query output an integer denoting maximum numbers of animals he could kill by shooting at a time during L and R (inclusive).

Constraints:

1<=N,Q<=100000

1<=X,Y,L,R<=1000000000

Example

Input:
4
1 2
2 3
4 5
6 7
4
1 5
2 3
4 7
5 7

Output:
2
2
1
1

 

一堆l和r超大的区间+1、查询区间max的操作

用个map离散一下完就是裸的线段树

技术分享 
 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
22     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
23     return x*f;
24 }
25 map<int,int>mp;
26 int n,m,cnt,cnt2;
27 int l[100010],r[100010];
28 int a[400010];
29 struct query{int l,r;}q[100010];
30 struct segtree{
31     int l,r,mx,tag;
32 }t[400010*4];
33 inline void pushdown(int k)
34 {int tt=t[k].tag;t[k].tag=0;t[k<<1].mx+=tt;t[k<<1].tag+=tt;t[k<<1|1].mx+=tt;t[k<<1|1].tag+=tt;}
35 inline void update(int k)
36 {t[k].mx=max(t[k<<1].mx,t[k<<1|1].mx);}
37 inline void buildtree(int now,int l,int r)
38 {
39     t[now].l=l;t[now].r=r;
40     if(l==r)
41     {
42         t[now].mx=t[now].tag=0;
43         return;
44     }
45     int mid=(l+r)>>1;
46     buildtree(now<<1,l,mid);
47     buildtree(now<<1|1,mid+1,r);
48 }
49 inline void add(int now,int x,int y,int d)
50 {
51     int l=t[now].l,r=t[now].r;
52     pushdown(now);
53     if (l==x&&r==y)
54     {
55         t[now].tag+=d;
56         t[now].mx+=d;
57         return;
58     }
59     int mid=(l+r)>>1;
60     if (y<=mid)add(now<<1,x,y,d);
61     else if (x>mid)add(now<<1|1,x,y,d);
62     else add(now<<1,x,mid,d),add(now<<1|1,mid+1,y,d);
63     update(now);
64 }
65 inline int ask(int now,int x,int y)
66 {
67     int l=t[now].l,r=t[now].r;
68     pushdown(now);
69     if (l==x&&r==y)return t[now].mx;
70     int mid=(l+r)>>1;
71     if (y<=mid)return ask(now<<1,x,y);
72     else if (x>mid)return ask(now<<1|1,x,y);
73     else return max(ask(now<<1,x,mid),ask(now<<1|1,mid+1,y));
74 }
75 int main()
76 {
77     n=read();
78     for (int i=1;i<=n;i++)l[i]=read(),r[i]=read(),a[++cnt]=l[i],a[++cnt]=r[i];
79     m=read();
80     for (int i=1;i<=m;i++)q[i].l=read(),q[i].r=read(),a[++cnt]=q[i].l,a[++cnt]=q[i].r;
81     sort(a+1,a+cnt+1);
82     for (int i=1;i<=cnt;i++)
83     if (i==1||a[i]!=a[i-1])mp[a[i]]=++cnt2;
84     for (int i=1;i<=n;i++)l[i]=mp[l[i]],r[i]=mp[r[i]];
85     for (int i=1;i<=m;i++)q[i].l=mp[q[i].l],q[i].r=mp[q[i].r];
86     buildtree(1,1,cnt2);
87     for (int i=1;i<=n;i++)add(1,l[i],r[i],1);
88     for (int i=1;i<=m;i++)printf("%d\n",ask(1,q[i].l,q[i].r));
89 }
Spoj BGSHOOT

 

Spoj-BGSHOOT


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