FJUT 3097（hdu 3333） 区间种类数 线段树+离线

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

区间种类数

TimeLimit:2500MS  MemoryLimit:128MB

64-bit integer IO format:%lld

已解决 | 点击收藏

Problem Description

给一长度为n的序列，询问其任意的区间的数字的种类数

Input

每个测试文件仅有一组数据
第一行是两个整数n和q代表，序列的长度和查询次数
接下来一行有n个数字 ，a1,a2,a3,……an 代表序列的n个元素
再接下有q行，每行有两个整数l,r,代表所要查询区间的左右端点

n<=100000
q<=100000
1<=l<=r<=n
1<=ai<=n

Output

输出区间[l,r]内数字的种类数

SampleInput

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

SampleOutput

2
2
3
5
6
5

很经典的问题，离线就是先按照右端点先排好序，对于出现过的a[1]来说，记录他上一次出现的位置。如果是第一次出现则记录这个位置并

在线段树上更新1，如果不是第一次，则把之前的位置更新为0，当前位置更新为1。线段树求区间和。

hdu3333同理。

/// .-~~~~~~~~~-._ _.-~~~~~~~~~-.
/// __.' ~. .~ `.__
/// .'// \./ \\`.
/// .'// | \\`.
/// .'// .-~"""""""~~~~-._ | _,-~~~~"""""""~-. \\`.
/// .'//.-" `-. | .-' "-.\\`.
/// .'//______.============-.. \ | / ..-============.______\\`.
/// .'______________________________\|/______________________________`.
#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_w
4000
ith_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=1e3+10;
const double EPS=1e-8;
const double eps=1e-8;
const int mod=1e9+7;
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;
//}

struct node
{
int lt, rt;
LL val;
}tree[4*maxn];

//向上更新
void PushUp(int id)
{
tree[id].val = tree[id<<1].val + tree[id<<1|1].val;
}

//建立线段树
void Build(int lt, int rt, int id)
{
tree[id].lt = lt;
tree[id].rt = rt;
tree[id].val = 0;//每段的初值，根据题目要求
if (lt == rt)
{
//tree[id].val = 1;
return;
}
int mid = (lt + rt) >> 1;
Build(lt, mid, id<<1);
Build(mid+1, rt, id<<1|1);
//PushUp(id);
}

//更改区间内某个点的值
void Change(int lt, int rt, int id, int to)
{
if (lt <= tree[id].lt && rt >= tree[id].rt)
{
tree[id].val = to;
return;
}
int mid = (tree[id].lt + tree[id].rt) >> 1;
if (lt <= mid)
Change(lt, rt, id<<1, to);
if (rt > mid)
Change(lt, rt, id<<1|1, to);
PushUp(id);
}

//查询某段区间内的he
LL Query(int lt, int rt, int id)
{
if (lt <= tree[id].lt && rt >= tree[id].rt)
return tree[id].val;
int mid = (tree[id].lt + tree[id].rt) >> 1;
LL ans = 0;
if (lt <= mid)
ans += Query(lt, rt, id<<1);
if (rt > mid)
ans += Query(lt, rt, id<<1|1);
return ans;
}

struct Node
{
int l,r;
int id;
}q[100005];

int a[100005],n,m;
LL sum[100005];

bool cmp(Node a,Node b)
{
return a.r<b.r;
}

void work()
{
Build(1,n,1);
int t,now=1;
map<int,int>s;
FOR(1,n,i)
{
t=s[a[i]];
if(t==0)
{
Change(i,i,1,1);//l r id value
//hdu 3333 Change(i,i,1,a[i]);
s[a[i]]=i;
}
else
{
Change(t,t,1,0);
Change(i,i,1,1);
//Change(i,i,1,a[i]);
s[a[i]]=i;
}
for(;now<=m&&q[now].r==i;now++)
sum[q[now].id]=Query(q[now].l,q[now].r,1);
}
}

void solve()
{
s_2(n,m);
FOR(1,n,i)
s_1(a[i]);
FOR(1,m,i)
{
s_2(q[i].l,q[i].r);
q[i].id=i;
}
sort(q+1,q+m+1,cmp);
work();
FOR(1,m,i)
print(sum[i]);
}
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();
}
}