FJUT 3097（hdu 3333） 区间种类数 主席树+在线

http://120.78.128.11/Problem.jsp?pid=3097

对每个位置建一棵主席树，每颗树记录的区间信息为[1~n]位置上的不重复权值和(第k大主席树记录的是值域，和这里不同)。从左往右建树，这样rt[i]树表示[1~i]所有不同值之和，然后和上面一样把重复出现的值的位置劲量往右移，当出现过a[i]时先删除rt[i-1]中对应位置的a[i]再把它加入到当前树中的i位置，查询的时候只要查询rt[r]树中[l,r]区间之和即可

/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#pragma GCC optimize(2)
#pragma comment(linker, "/STACK:102400000,102400000")
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
#include <stdlib.h>
#include <time.h>
#include <bitset>
using namespace std;

#define pi acos(-1)
#define s_1(x) scanf("%d",&x)
#define s_2(x,y) scanf("%d%d",&x,&y)
#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)
#define S_1(x) scan_d(x)
#define S_2(x,y) scan_d(x),scan_d(y)
#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)
#define PI acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x,y) memset(x,y,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define fOR(n,x,i) for(int i=n;i>=x;i--)
#define fOr(n,x,i) for(int i=n;i>x;i--)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)
#define bug printf("***********\n");
#define db double
#define ll long long
#define mp make_pair
#define pb push_back
typedef long long LL;
typedef pair <int, int> ii;
const int INF=~0U>>1;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
const int dx[]={-1,0,1,0,1,-1,-1,1};
const int dy[]={0,1,0,-1,-1,1,-1,1};
const int maxn=1e5+10;
const int maxx=1e6+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=19260817;
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
template <class T>
inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}
while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}

inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;
while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}
else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}
if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}
if(IsN) num=-num;return true;}

void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}
void print(LL a){ Out(a),puts("");}
//freopen( "in.txt" , "r" , stdin );
//freopen( "data.txt" , "w" , stdout );
//cerr << "run time is " << clock() << endl;

//void readString(string &s)
//{
// static char str[maxn];
// scanf("%s", str);
// s = str;
//}
const int MAXM=maxn*40;
int lc[MAXM],rc[MAXM],tol;
LL sum[MAXM];
int a[maxn], rt[maxn];
map<int,int>vis;

int n,q;
void init()
{
tol=1;
vis.clear();
}
void update(int &u, int x, int y, int p, LL v)
{
sum[tol] = sum[u]+v,lc[tol] = lc[u],rc[tol] = rc[u];
u = tol++;
if(x == y) return;
int m = (x+y)>>1;
if(p <= m) update(lc[u],x,m,p,v);
else update(rc[u],m+1,y,p,v);
}
LL query(int u, int x, int y, int ql, int qr)
{
if(ql <= x && y <= qr) return sum[u];
int m = (x+y)>>1;
LL ans = 0;
if(ql <= m) ans += query(lc[u],x,m,ql,qr);
if(qr > m) ans += query(rc[u],m+1,y,ql,qr);
return ans;
}

void solve()
{
s_2(n,q);
init();
FOR(1,n,i) s_1(a[i]);
FOR(1,n,i)
{
rt[i]=rt[i-1];
if(vis.find(a[i])!=vis.end())
{
int tmp=rt[i-1];
update(tmp,1,n,vis[a[i]],-1);
rt[i]=tmp;
update(rt[i],1,n,i,1);
}
else
{
rt[i] = rt[i-1];
update(rt[i],1,n,i,1);
}
vis[a[i]]=i;
}
W(q--)
{
int l,r;
s_2(l,r);
printf("%lld\n", query(rt[r],1,n,l,r));
}
}

int main()
{
//freopen( "1.in" , "r" , stdin );
//freopen( "1.out" , "w" , stdout );
int t=1;
//init();
//s_1(t);
for(int cas=1;cas<=t;cas++)
{
//printf("Case #%d: ",cas);
solve();
}
}