SPOJ DQUERY D-query (主席树)

题意：给定上一个序列，然后有一些询问，问你区间 l - r 中有多少个不同的数。

析：一个主席树入门题，首先是先进行处理，记录不同数出现的个数，如果相同的，先减去以前的，再加上这个最新的，

对于查询，处理好，每一部分。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 30000 + 10;
const int maxm = maxn * 100;
const int mod = 10;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn], T[maxn];
int lch[maxm], rch[maxm], c[maxm];
int tot;
map<int, int> mp;

int build(int l, int r){
  int rt = tot++;
  c[rt] = 0;
  if(l == r)  return rt;
  int m = l + r >> 1;
  lch[rt] = build(l, m);
  rch[rt] = build(m+1, r);
  return rt;
}

int update(int rt, int pos, int val){
  int newrt = tot++;
  int tmp = newrt;
  int l = 1, r = n;
  c[newrt] = c[rt] + val;
  while(l < r){
    int m = l + r >> 1;
    if(pos <= m){
      lch[newrt] = tot++;
      rch[newrt] = rch[rt];
      newrt = lch[newrt];
      rt = lch[rt];
      r = m;
    }
    else{
      rch[newrt] = tot++;
      lch[newrt] = lch[rt];
      newrt = rch[newrt];
      rt = rch[rt];
      l = m + 1;
    }
    c[newrt] = c[rt] + val;
  }
  return tmp;
}

int query(int rt, int pos){
  int l = 1, r = n;
  int ans = 0;
  while(pos < r){
    int m = l + r >> 1;
    if(pos <= m){
      rt = lch[rt];
      r = m;
    }
    else{
      l = m + 1;
      ans += c[lch[rt]];
      rt = rch[rt];
    }
  }
  return ans + c[rt];
}

int main(){
  while(scanf("%d", &n) == 1){
    for(int i = 1; i <= n; ++i)
      scanf("%d", a+i);
    mp.clear();
    tot = 0;
    T[n+1] = build(1, n);
    for(int i = n; i; --i){
      if(mp.count(a[i])){
        int tmp = update(T[i+1], mp[a[i]], -1);
        T[i] = update(tmp, i, 1);
      }
      else T[i] = update(T[i+1], i, 1);
      mp[a[i]] = i;
    }
    scanf("%d", &m);
    while(m--){
      int l, r;
      scanf("%d %d", &l, &r);
      printf("%d
", query(T[l], r));
    }
  }
  return 0;
}