spoj DQUERY - D-query (询问区间中不同数的个数，线段树(离线) or 莫队算法（离线） or 主席树（在线）)
题目链接 题意：给出n个数，然后m个询问，每个询问一个区间[l,r],问该区间中不同的数有多少个。
思路：离线处理+线段树的做法不多说了：
1/* ***********************************************
2Author :111qqz
3Created Time :Fri 16 Sep 2016 11:34:32 PM CST
4File Name :code/spoj/dquery.cpp
5************************************************ */
6#include <cstdio>
7#include <cstring>
8#include <iostream>
9#include <algorithm>
10#include <vector>
11#include <queue>
12#include <set>
13#include <map>
14#include <string>
15#include <cmath>
16#include <cstdlib>
17#include <ctime>
18#define fst first
19#define sec second
20#define lson l,m,rt<<1
21#define rson m+1,r,rt<<1|1
22#define ms(a,x) memset(a,x,sizeof(a))
23typedef long long LL;
24#define pi pair < int ,int >
25#define MP make_pair
26using namespace std;
27const double eps = 1E-8;
28const int dx4[4]={1,0,0,-1};
29const int dy4[4]={0,-1,1,0};
30const int inf = 0x3f3f3f3f;
31const int N=3E4+7;
32const int M=2E5+7;
33int n,Q;
34int a[N];
35int tree[N<<2];
36map<int,int>mp;
37struct node
38{
39 int l,r;
40 int id;
41 bool operator < (node b)const
42 {
43 if (r==b.r) return l<b.l;
44 return r<b.r;
45 }
46}q[M];
47void PushUp( int rt)
48{
49 tree[rt] = tree[rt<<1] + tree[rt<<1|1];
50}
51void update( int p,int sc,int l,int r ,int rt)
52{
53 if (l==r)
54 {
55 tree[rt]+=sc;
56 return;
57 }
58 int m = (l+r)>>1;
59 if (p<=m) update(p,sc,lson);
60 else update(p,sc,rson);
61 PushUp(rt);
62}
63int query(int L,int R,int l,int r,int rt)
64{
65 if (L<=l&&r<=R) return tree[rt];
66 int m = (l+r)>>1;
67 int ret = 0 ;
68 if (L<=m) ret += query(L,R,lson);
69 if (R>=m+1) ret+=query(L,R,rson);
70 return ret;
71}
72int ans[M];
73int main()
74{
75 #ifndef ONLINE_JUDGE
76 freopen("code/in.txt","r",stdin);
77 #endif
78 cin>>n;
79 for ( int i = 1 ; i <= n ; i++) scanf("%d",&a[i]);
80 cin>>Q;
81 for ( int i = 1 ; i <= Q ; i++) scanf("%d %d",&q[i].l,&q[i].r),q[i].id = i ;
82 sort(q+1,q+Q+1);
83 int cur = 1;
84 for ( int i = 1 ; i <= Q ; i++)
85 {
86 for ( ; cur <= q[i].r ; cur++)
87 {
88 if (mp[a[cur]]) update(mp[a[cur]],-1,1,n,1);
89 mp[a[cur]] = cur;
90 update(mp[a[cur]],1,1,n,1);
91 }
92 ans[q[i].id] = query(q[i].l,q[i].r,1,n,1);
93 }
94 for ( int i = 1 ; i <= Q ; i++) printf("%d\n",ans[i]);
95 #ifndef ONLINE_JUDGE
96 fclose(stdin);
97 #endif
98 return 0;
99}
之后补一个主席树的做法
先来补一个莫队的做法。
因为a[i]<=1E6,可以很方便得统计每个数出现的次数，update的时候，如果是1变成0，那么区间中不同的数的个数减1，如果是0变成1，那么区间中不同数个数+1
/* ***********************************************
Author :111qqz
Created Time :Fri 16 Sep 2016 11:34:32 PM CST
File Name :code/spoj/dquery.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int MAXN = 1E6+7;
7const int N=3E4+7;
8const int M=2E5+7;
9int n,Q;
10int a[N];
11int pos[N];
12int sum;
13int cnt[MAXN]={0};
14struct node
15{
16 int l,r;
17 int id;
18 bool operator < (node b)const
19 {
20 if (pos[l]==pos[b.l]) return r<b.r;
21 return pos[l] < pos[b.l];
22 }
23}q[M];
1int ans[M];
2void update ( int x,int d)
3{
4 cnt[a[x]]+=d;
5 if (d==1&&cnt[a[x]]==1) sum++;
6 if (d==-1&&cnt[a[x]]==0) sum--;
7}
1int main()
2{
3 #ifndef ONLINE_JUDGE
4 freopen("./in.txt","r",stdin);
5 #endif
6 int bk = 173;// sqrt(30000)
7 cin>>n;
8 for ( int i = 1 ; i <= n ; i++)
9 {
10 scanf("%d",&a[i]);
11 pos[i] = (i-1)/bk;
12 }
13 cin>>Q;
14 for ( int i = 1 ; i <= Q ; i++) scanf("%d %d",&q[i].l,&q[i].r),q[i].id = i ;
15 sort(q+1,q+Q+1);
16 int pl=1,pr=0,id,l,r;
17 ms(cnt,0);
18 sum = 0 ;
19 for ( int i = 1 ; i <= Q ; i++)
20 {
21 id = q[i].id;
22 l = q[i].l;
23 r = q[i].r;
24 if (pr<r)
25 {
26 for ( int j = pr+1 ; j <= r ; j++)
27 update(j,1);
28 }
29 else
30 {
31 for ( int j = r+1 ; j <= pr ; j++)
32 update(j,-1);
33 }
34 pr = r;
35 if (pl<l)
36 {
37 for ( int j = pl ; j <= l-1 ; j++)
38 update(j,-1);
39 }
40 else
41 {
42 for ( int j = l ; j <= pl-1 ; j++)
43 update(j,1);
44 }
45 pl = l;
46 ans[id] = sum;
47 }
48 for ( int i = 1 ; i <= Q ; i++) printf("%d\n",ans[i]);
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}
补一个在线的做法，可持久化线段树，其实思路和离线线段树几乎一样。
/* ***********************************************
Author :111qqz
Created Time :Fri 16 Sep 2016 11:34:32 PM CST
File Name :code/spoj/dquery.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=3E4+7;
7const int M=2E5+7;
8int n,Q;
9int a[N],root[N];
10map<int,int>mp;
11int cnt;
12struct Ptree
13{
14 int sum;
15 int left,right;
16}tree[N*30];
17void build (int &rt,int l,int r)
18{
19 rt =++cnt;
20 tree[rt].sum = 0 ;
21 if (l==r) return;
22 int mid = (l+r)>>1;
23 build (tree[rt].left,l,mid);
24 build (tree[rt].right,mid+1,r);
25}
26inline int add_node(int _sum,int _left,int _right)
27{
28 int idx = ++cnt;
29 tree[idx].sum = _sum;
30 tree[idx].left = _left;
31 tree[idx].right = _right;
32 return idx;
33}
34void Insert(int &root,int pre_rt,int pos,int val,int l,int r)
35{
36// printf("l:%d r:%d\n",l,r);
37 root = add_node(tree[pre_rt].sum + val,tree[pre_rt].left,tree[pre_rt].right);
38 // root = ++cnt;
39 // tree[root].sum = tree[pre_rt].sum+ val;
40 // tree[root].left = tree[pre_rt].left;
41 // tree[root].right = tree[pre_rt].right;
42 if (l==r) return;
43 int mid = (l+r)>>1;
44 if (pos<=mid) Insert(tree[root].left,tree[pre_rt].left,pos,val,l,mid);
45 else Insert(tree[root].right,tree[pre_rt].right,pos,val,mid+1,r);
46}
47int query(int pos,int l,int r,int rt)
48{
49 if (l==r) return tree[rt].sum;
50 int mid = (l+r)>>1;
51 if (pos<=mid) return query(pos,l,mid,tree[rt].left);
52 else return tree[tree[rt].left].sum + query(pos,mid+1,r,tree[rt].right);
53}
54int main()
55{
56 #ifndef ONLINE_JUDGE
57 freopen("./in.txt","r",stdin);
58 #endif
59 ms(tree,0);
60 cnt = 0 ;
61 cin>>n;
62 for ( int i = 1 ; i <= n ; i++) scanf("%d",&a[i]);
63 build (root[n+1],1,n);
64 mp.clear();
65 for ( int i = n ; i >= 1 ; i--)
66 {
67 if (mp.find(a[i])==mp.end())
68 {
69 Insert(root[i],root[i+1],i,1,1,n);
70 }
71 else
72 {
73 Insert(root[i],root[i+1],mp[a[i]],-1,1,n);
74 Insert(root[i],root[i],i,1,1,n);
75 }
76 mp[a[i]] = i;
77 }
78 cin>>Q;
79 while (Q--)
80 {
81 int l,r;
82 scanf("%d %d",&l,&r);
83 // printf("l:%d r:%d\n",l,r);
84 int ans = query(r,1,n,root[l]);
85 printf("%d\n",ans);
86 }
87 #ifndef ONLINE_JUDGE
88 fclose(stdin);
89 #endif
90 return 0;
91}