hdu 5919 Sequence II (主席树，求区间不同数的个数)

题意：

　　给定一个序列

nn

，有

mm

次查询，每次查询一个区间

[l,r][l,r]

，求区间中每一种数在区间中第一次出现的位置的中位数，强制在线。

题解：

　　强制在线

　　利用主席树求区间不同数的个数

　　这里有个技巧

　　倒着插入主席树
　　在寻找位置的中位数上就可以一个log解决了
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 1000100
#define MOD 1000000007

const int MAXN = 200100;
const int M = MAXN * 50;
int n,m,tot;
int a[MAXN];
int T[MAXN],lson[M],rson[M],c[M];
int build(int l,int r)
{
int root = tot++;
c[root] = 0;
if(l != r)
{
int mid = (l+r)>>1;
lson[root] = build(l,mid);
rson[root] = build(mid+1,r);
}
return root;
}
int update(int root,int pos,int val)
{
int newroot = tot++,tmp = newroot;
c[newroot] = c[root] + val;
int l = 1,r = n;
while(l < r)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
lson[newroot] = tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
rson[newroot] = tot++;
lson[newroot] = lson[root];
newroot = rson[newroot];
root = rson[root];
l = mid + 1;
}
c[newroot] = c[root] + val;
}
return tmp;
}
int query(int root,int pos)
{
int ret = 0;
int l = 1,r = n;
while(pos < r)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
r = mid;
root = lson[root];
}
else
{
ret += c[lson[root]];
root = rson[root];
l = mid + 1;
}
}
return ret + c[root];
}
int solve(int x,int l,int r,int k)
{
if(l == r)
return l;
int mid = (l+r)>>1;
if(c[lson[x]] >= k)
return solve(lson[x],l,mid,k);
return solve(rson[x],mid + 1,r,k-c[lson[x]]);
}
int last[MAXN];
int main()
{
int t,C = 1;
scanf("%d",&t);
while(t--)
{
tot = 0;
scanf("%d%d",&n,&m);
for(int i = 1; i <= n; i++)
scanf("%d",&a[i]);
memset(c,0,sizeof(c));
memset(lson,0,sizeof(lson));
memset(rson,0,sizeof(rson));
T[n+1] = build(1,n);
memset(last,-1,sizeof(last));
for(int i = n; i >= 1; i--)
{
if(last[a[i]] == -1)
T[i] = update(T[i+1],i,1);
else
{
int tmp = update(T[i+1],last[a[i]],-1);
T[i] = update(tmp,i,1);
}
last[a[i]] = i;
}
int pre = 0;
printf("Case #%d:",C++);
for(int i = 0; i < m; i++)
{
int l,r,L,R;
scanf("%d%d",&L,&R);
l = min((L+pre)%n+1,(R+pre)%n+1);
r = max((L+pre)%n+1,(R+pre)%n+1);
int k = (query(T[l],r) + 1) >> 1;
pre = solve(T[l],1,n,k);
printf(" %d",pre);
}
printf("\n");
}
return 0;
}