Bzoj3524: [poi2014] couriers

3524: [poi2014] courierstime limit: 20 sec memory limit: 128 MB
Submit: 466 solved: 141
[Submit] [Status] Description

Give a sequence a with the length of N. 1 ≤ A [I] ≤ n.
The M group asks if a [L, R] interval exists and whether a number appears in [L, R] More than (R-l + 1)/2. If yes, this number is output; otherwise, 0 is output.

Input

The first row has two numbers N and M.
Number of N in the second row, a [I].
In the next m row, there are two numbers in each row, L and R, indicating to ask about the range [L, R.

Output

M rows. Each row corresponds to one answer.

Sample input7 5
1 1 3 2 3 4 3
1 3
1 4
3 7
1 7
6 6

Sample output1
0
3
0
4

Hint

[Data Scope]

N, m ≤ 500000

Source

By dzy

Question:

The Chairman tree bug also exists...

Because the range Subtraction is satisfied, we can obtain the weight information of the range [L, R] In a tree-like array, and then this question is bare...

Code:

  1 #include<cstdio>  2   3 #include<cstdlib>  4   5 #include<cmath>  6   7 #include<cstring>  8   9 #include<algorithm> 10  11 #include<iostream> 12  13 #include<vector> 14  15 #include<map> 16  17 #include<set> 18  19 #include<queue> 20  21 #include<string> 22  23 #define inf 1000000000 24  25 #define maxn 10000000+1000 26  27 #define maxm 500000+1000 28  29 #define eps 1e-10 30  31 #define ll long long 32  33 #define pa pair<int,int> 34  35 #define for0(i,n) for(int i=0;i<=(n);i++) 36  37 #define for1(i,n) for(int i=1;i<=(n);i++) 38  39 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 40  41 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 42  43 #define mod 1000000007 44  45 using namespace std; 46  47 inline int read() 48  49 { 50  51     int x=0,f=1;char ch=getchar(); 52  53     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 54  55     while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();} 56  57     return x*f; 58  59 } 60 int n,m,tot,root[maxm],ls[maxn],rs[maxn],s[maxn]; 61 void update(int l,int r,int x,int &y,int z) 62 { 63     y=++tot; 64     s[y]=s[x]+1; 65     if(l==r)return; 66     ls[y]=ls[x];rs[y]=rs[x]; 67     int mid=(l+r)>>1; 68     if(z<=mid)update(l,mid,ls[x],ls[y],z);else update(mid+1,r,rs[x],rs[y],z); 69 } 70 int query(int x,int y) 71 { 72     int l=1,r=n,mid,xx=root[x-1],yy=root[y],tmp=(y-x+1)>>1; 73     while(l!=r) 74     { 75         mid=(l+r)>>1; 76         if(s[yy]-s[xx]<tmp)return 0; 77         if(s[ls[yy]]-s[ls[xx]]>tmp){r=mid;xx=ls[xx];yy=ls[yy];} 78         else if(s[rs[yy]]-s[rs[xx]]>tmp){l=mid+1;xx=rs[xx];yy=rs[yy];} 79         else return 0; 80     } 81     return l; 82 } 83  84 int main() 85  86 { 87  88     freopen("input.txt","r",stdin); 89  90     freopen("output.txt","w",stdout); 91  92     n=read();m=read(); 93     for1(i,n) 94     { 95         int x=read();update(1,n,root[i-1],root[i],x); 96     } 97     while(m--) 98     { 99         int x=read(),y=read();100         printf("%d\n",query(x,y));101     }102 103     return 0;104 105 }

View code

 

Bzoj3524: [poi2014] couriers