HDU-3874 Necklace 线段树+离线

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3874

　　比较简单的题，题意也好懂。

　　先O(n)求每个数左边第一次出现的与他相同的数的位置l[i]。对询问按照y从小大排序，然后按照从左到右的顺序来跟新点，当前点为i，那么删掉l[i]，加入点i，然后遇到询问求和。

 //STATUS:C++_AC_2593MS_10024KB
 #include <functional>
 #include <algorithm>
 #include <iostream>
 //#include <ext/rope>
 #include <fstream>
 #include <sstream>
 #include <iomanip>
 #include <numeric>
 #include <cstring>
 #include <cassert>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <bitset>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,102400000")
 //using namespace __gnu_cxx;
 //define
 #define pii pair<int,int>
 #define mem(a,b) memset(a,b,sizeof(a))
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define PI acos(-1.0)
 //typedef
 typedef __int64 LL;
 typedef unsigned __int64 ULL;
 //const
 const int N=,M=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //Daily Use ...
 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 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));}
 //End

 struct Node{
     int a,b,id;
     bool operator < (const Node& a)const {
         return b<a.b;
     }
 }q[M];
 LL sum[N<<],ans[M];
 int num[N],l[N],la[];
 int T,n,m;

 void update(int l,int r,int rt,int w,int val)
 {
     if(l==r){
         sum[rt]=val;
         return;
     }
     int mid=(l+r)>>;
     if(w<=mid)update(lson,w,val);
     else update(rson,w,val);
     sum[rt]=sum[rt<<]+sum[rt<<|];
 }

 LL query(int l,int r,int rt,int L,int R)
 {
     if(L<=l && r<=R){
         return sum[rt];
     }
     int mid=(l+r)>>;
     LL ret=;
     if(L<=mid)ret+=query(lson,L,R);
     if(R>mid)ret+=query(rson,L,R);
     return ret;
 }

 int main()
 {
   //  freopen("in.txt","r",stdin);
     int i,j,k;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         mem(la,);
         for(i=;i<=n;i++){
             scanf("%d",&num[i]);
             l[i]=la[num[i]];
             la[num[i]]=i;
         }
         scanf("%d",&m);
         for(i=;i<m;i++){
             scanf("%d%d",&q[i].a,&q[i].b);
             q[i].id=i;
         }
         sort(q,q+m);
         mem(sum,);k=;
         for(i=;i<=n;i++){
             if(l[i]){
                 update(,n,,l[i],);
             }
             update(,n,,i,num[i]);
             for(;q[k].b==i && k<m;k++){
                 ans[q[k].id]=query(,n,,q[k].a,q[k].b);
             }
         }
         for(i=;i<m;i++){
             printf("%I64d\n",ans[i]);
         }
     }
     return ;
 }