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
1/* ***********************************************
2Author :111qqz
3Created Time :Fri 16 Sep 2016 11:34:32 PM CST
4File Name :code/spoj/dquery.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int MAXN = 1E6+7;
34const int N=3E4+7;
35const int M=2E5+7;
36int n,Q;
37int a[N];
38int pos[N];
39int sum;
40int cnt[MAXN]={0};
41struct node
42{
43 int l,r;
44 int id;
45 bool operator < (node b)const
46 {
47 if (pos[l]==pos[b.l]) return r<b.r;
48 return pos[l] < pos[b.l];
49 }
50}q[M];
51
52int ans[M];
53void update ( int x,int d)
54{
55 cnt[a[x]]+=d;
56 if (d==1&&cnt[a[x]]==1) sum++;
57 if (d==-1&&cnt[a[x]]==0) sum--;
58}
59
60int main()
61{
62 #ifndef ONLINE_JUDGE
63 freopen("./in.txt","r",stdin);
64 #endif
65 int bk = 173;// sqrt(30000)
66 cin>>n;
67 for ( int i = 1 ; i <= n ; i++)
68 {
69 scanf("%d",&a[i]);
70 pos[i] = (i-1)/bk;
71 }
72 cin>>Q;
73 for ( int i = 1 ; i <= Q ; i++) scanf("%d %d",&q[i].l,&q[i].r),q[i].id = i ;
74 sort(q+1,q+Q+1);
75 int pl=1,pr=0,id,l,r;
76 ms(cnt,0);
77 sum = 0 ;
78 for ( int i = 1 ; i <= Q ; i++)
79 {
80 id = q[i].id;
81 l = q[i].l;
82 r = q[i].r;
83 if (pr<r)
84 {
85 for ( int j = pr+1 ; j <= r ; j++)
86 update(j,1);
87 }
88 else
89 {
90 for ( int j = r+1 ; j <= pr ; j++)
91 update(j,-1);
92 }
93 pr = r;
94 if (pl<l)
95 {
96 for ( int j = pl ; j <= l-1 ; j++)
97 update(j,-1);
98 }
99 else
100 {
101 for ( int j = l ; j <= pl-1 ; j++)
102 update(j,1);
103 }
104 pl = l;
105 ans[id] = sum;
106 }
107 for ( int i = 1 ; i <= Q ; i++) printf("%d\n",ans[i]);
108
109
110 #ifndef ONLINE_JUDGE
111 fclose(stdin);
112 #endif
113 return 0;
114}
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。
1/* ***********************************************
2Author :111qqz
3Created Time :Fri 16 Sep 2016 11:34:32 PM CST
4File Name :code/spoj/dquery.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=3E4+7;
34const int M=2E5+7;
35int n,Q;
36int a[N],root[N];
37map<int,int>mp;
38int cnt;
39struct Ptree
40{
41 int sum;
42 int left,right;
43}tree[N*30];
44void build (int &rt,int l,int r)
45{
46 rt =++cnt;
47 tree[rt].sum = 0 ;
48 if (l==r) return;
49 int mid = (l+r)>>1;
50 build (tree[rt].left,l,mid);
51 build (tree[rt].right,mid+1,r);
52}
53inline int add_node(int _sum,int _left,int _right)
54{
55 int idx = ++cnt;
56 tree[idx].sum = _sum;
57 tree[idx].left = _left;
58 tree[idx].right = _right;
59 return idx;
60}
61void Insert(int &root,int pre_rt,int pos,int val,int l,int r)
62{
63// printf("l:%d r:%d\n",l,r);
64 root = add_node(tree[pre_rt].sum + val,tree[pre_rt].left,tree[pre_rt].right);
65 // root = ++cnt;
66 // tree[root].sum = tree[pre_rt].sum+ val;
67 // tree[root].left = tree[pre_rt].left;
68 // tree[root].right = tree[pre_rt].right;
69 if (l==r) return;
70 int mid = (l+r)>>1;
71 if (pos<=mid) Insert(tree[root].left,tree[pre_rt].left,pos,val,l,mid);
72 else Insert(tree[root].right,tree[pre_rt].right,pos,val,mid+1,r);
73}
74int query(int pos,int l,int r,int rt)
75{
76 if (l==r) return tree[rt].sum;
77 int mid = (l+r)>>1;
78 if (pos<=mid) return query(pos,l,mid,tree[rt].left);
79 else return tree[tree[rt].left].sum + query(pos,mid+1,r,tree[rt].right);
80}
81int main()
82{
83 #ifndef ONLINE_JUDGE
84 freopen("./in.txt","r",stdin);
85 #endif
86 ms(tree,0);
87 cnt = 0 ;
88 cin>>n;
89 for ( int i = 1 ; i <= n ; i++) scanf("%d",&a[i]);
90 build (root[n+1],1,n);
91 mp.clear();
92 for ( int i = n ; i >= 1 ; i--)
93 {
94 if (mp.find(a[i])==mp.end())
95 {
96 Insert(root[i],root[i+1],i,1,1,n);
97 }
98 else
99 {
100 Insert(root[i],root[i+1],mp[a[i]],-1,1,n);
101 Insert(root[i],root[i],i,1,1,n);
102 }
103 mp[a[i]] = i;
104 }
105 cin>>Q;
106 while (Q--)
107 {
108 int l,r;
109 scanf("%d %d",&l,&r);
110 // printf("l:%d r:%d\n",l,r);
111 int ans = query(r,1,n,root[l]);
112 printf("%d\n",ans);
113 }
114 #ifndef ONLINE_JUDGE
115 fclose(stdin);
116 #endif
117 return 0;
118}