[BZOJ 1878][SDOI2009]HH的项链 (离线,树状数组)

分析看这个大神的，我的还是参考他说的算法实现的：http://www.dxmtb.com/blog/diff/ 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxm = 1000005;
const int maxn = 50005;


int C[maxn],next[maxn],B[maxn];
int flag[maxm];
struct query{
    int l,r;
    int qnum,ans;
}Q[200005]; 
int N,M;
int cmp1(query a,query b){
    if(a.l == b.l) return a.r < b.r;
    else           return a.l < b.l;
}
int cmp2(query a,query b){
    return a.qnum < b.qnum;
}
int lowbit(int x){
    return x&(-x);
}
int sum(int x){
    int temp = 0;
    while(x >= 1){
        temp += C[x];
        x -= lowbit(x);
    }
    return temp;
}
void update(int x,int num){
    while(x <= N){
        C[x] += num;
        x += lowbit(x);
    }
    return;
}
int main()
{
    //if(freopen("input.txt","r",stdin)== NULL)  {printf("Error\n"); exit(0);}
    cin>>N;
    memset(flag,-1,sizeof(flag));
    memset(C,0,sizeof(C));
    memset(B,0,sizeof(B));
    memset(next,0,sizeof(next));
       for(int i=1;i<=N;i++){
           int a;
           scanf("%d",&a);
           if(flag[a]== -1){
               flag[a] = i;  
               B[i] = 1;
           }
           else{
               next[flag[a]] = i;
               flag[a] = i;
           }
    }
    for(int i=1;i<=N;i++)
      for(int j=i-lowbit(i)+1;j<=i;j++){
           C[i] += B[j]; 
      }
    cin>>M;
    for(int i=1;i<=M;i++){
        scanf("%d%d",&Q[i].l,&Q[i].r);
        Q[i].qnum = i;
    }
    sort(Q+1,Q+M+1,cmp1); 
    int pv = 1;
    for(int i=1;i<=N;i++){
        while(Q[pv].l == i){
            Q[pv].ans = sum(Q[pv].r) - sum(Q[pv].l -1);
                       //printf("%d %d %d %d\n",Q[pv].r,sum(Q[pv].r) ,Q[pv].l -1,sum(Q[pv].l -1));
            pv++;   //for(int i=1;i<=N;i++) printf("%d ",C[i]); printf("\n");
        }
        if(next[i]) {
            update(next[i],1);
            update(i,-1);
        }
        
    }
    sort(Q+1,Q+M+1,cmp2);
    for(int i=1;i<=M;i++){
        printf("%d\n",Q[i].ans);
    }
    return 0;
}