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=50010,M=200010;
const int INF=0x3f3f3f3f;
const int MOD=95041567,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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<<2],ans[M];
int num[N],l[N],la[1000010];
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)>>1;
    if(w<=mid)update(lson,w,val);
    else update(rson,w,val);
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

LL query(int l,int r,int rt,int L,int R)
{
    if(L<=l && r<=R){
        return sum[rt];
    }
    int mid=(l+r)>>1;
    LL ret=0;
    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,0);
        for(i=1;i<=n;i++){
            scanf("%d",&num[i]);
            l[i]=la[num[i]];
            la[num[i]]=i;
        }
        scanf("%d",&m);
        for(i=0;i<m;i++){
            scanf("%d%d",&q[i].a,&q[i].b);
            q[i].id=i;
        }
        sort(q,q+m);
        mem(sum,0);k=0;
        for(i=1;i<=n;i++){
            if(l[i]){
                update(1,n,1,l[i],0);
            }
            update(1,n,1,i,num[i]);
            for(;q[k].b==i && k<m;k++){
                ans[q[k].id]=query(1,n,1,q[k].a,q[k].b);
            }
        }
        for(i=0;i<m;i++){
            printf("%I64d
",ans[i]);
        }
    }
    return 0;
}