bzoj 2223 [Coci 2009]PATULJCI

[Coci 2009]PATULJCI

Time Limit: 10 Sec Memory Limit: 259 MB
Submit: 1286 Solved: 553
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Description

Input

Output

10 3
1 2 1 2 1 2 3 2 3 3
8
1 2
1 3
1 4
1 5
2 5
2 6
6 9
7 10

Sample Input

no
yes 1
no
yes 1
no
yes 2
no
yes 3

Sample Output

HINT

Notice:输入第二个整数是序列中权值的范围Lim，即1<=ai(1<=i<=n)<=Lim。

1<=Lim<=10000

题解：

　　这道题就是可持久化线段树吧，然后维护一下该节点的size即可，权值线段树。

 1 #include<cstring>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<cstdio>
 6 
 7 #define N 300007
 8 #define M 10000007
 9 using namespace std;
10 inline int read()
11 {
12     int x=0,f=1;char ch=getchar();
13     while(ch>'9'||ch<'0'){if (ch=='-') f=-1;ch=getchar();}
14     while(ch<='9'&&ch>='0')
15     {
16         x=(x<<3)+(x<<1)+ch-'0';
17         ch=getchar();
18     }
19     return x*f;
20 }
21 
22 int n,m,sz,limit;
23 int root[N],ls[M],rs[M],sum[M];
24 
25 void change(int l,int r,int yl,int &xz,int z)
26 {
27     xz=++sz,sum[xz]=sum[yl]+1;
28     if (l==r) return;
29     ls[xz]=ls[yl],rs[xz]=rs[yl];
30     int mid=(l+r)>>1;
31     if (z<=mid) change(l,mid,ls[yl],ls[xz],z);
32     else change(mid+1,r,rs[yl],rs[xz],z);
33 }
34 int query(int l,int r,int x,int y,int tmp)
35 {
36     if (l==r) return l;
37     int mid=(l+r)>>1;
38     if (sum[ls[y]]-sum[ls[x]]>tmp) return query(l,mid,ls[x],ls[y],tmp);
39     else if (sum[rs[y]]-sum[rs[x]]>tmp) return query(mid+1,r,rs[x],rs[y],tmp);
40     else return 0;
41 }
42 int main()
43 {
44     n=read(),limit=read();
45     for (int i=1;i<=n;i++)
46     {
47         int x=read();
48         change(1,limit,root[i-1],root[i],x);
49     }
50     m=read();
51     for (int i=1;i<=m;i++)
52     {
53         int x=read(),y=read();
54         int ans=query(1,limit,root[x-1],root[y],(y-x+1)/2);
55         if (ans==0) printf("no
");
56         else printf("yes %d
",ans);
57     }
58 }