hdu 5654 xiaoxin and his watermelon candy 树状数组维护区间唯一元组

题目链接

题意:序列长度为n(1<= n <= 200,000)的序列，有Q(<=200,000)次区间查询，问区间[l,r]中有多少个不同的连续递增的三元组。

思路:连续三元组->递推O(n)将第一次出现该三元组的下标记录到树状数组中，并且用一个Next[]来表示递推关系，即同一个三元组下一次出现的位置是Next[x];这很关键，在之后按照左边界处理时(有点像莫队算法)，一直递推在l+1左侧重复出现的三元组，为了把该三元组推到出现在[l,r]或者[r+1,..]中，这样不重不漏。

使用map维护出现的id即可，题目很经典~~

#include<bits/stdc++.h>
using namespace std;
#define rep0(i,l,r) for(int i = (l);i < (r);i++)
#define rep1(i,l,r) for(int i = (l);i <= (r);i++)
#define rep_0(i,r,l) for(int i = (r);i > (l);i--)
#define rep_1(i,r,l) for(int i = (r);i >= (l);i--)
#define MS0(a) memset(a,0,sizeof(a))
#define MS1(a) memset(a,-1,sizeof(a))
#define MSi(a) memset(a,0x3f,sizeof(a))
#define inf 0x3f3f3f3f
#define lson l, m, rt << 1
#define rson m+1, r, rt << 1|1
typedef pair<int,int> PII;
#define A first
#define B second
#define MK make_pair
typedef __int64 ll;
typedef unsigned int uint;
template<typename T>
void read1(T &m)
{
    T x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    m = x*f;
}
template<typename T>
void read2(T &a,T &b){read1(a);read1(b);}
template<typename T>
void read3(T &a,T &b,T &c){read1(a);read1(b);read1(c);}
template<typename T>
void out(T a)
{
    if(a>) out(a/);
    putchar(a%+'');
}
int T,kase = ,i,j,k,n,m,l,r;
#define N 200200
map<pair<PII,int>,int> mp;
int a[N],B[N],Next[N],x[N],y[N],o[N],ans[N];
bool cmp(int a,int b){return x[a] < x[b];}
#define lowbit(x) (x&(-x))
void update(int p,int d)
{
    while(p <= n) B[p] += d,p += lowbit(p);
}
int query(int x)
{
    int ans = ;
    while(x) ans += B[x],x -= lowbit(x);
    return ans;
}
int main()
{
    //freopen("data.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    read1(T);
    while(T--){
        mp.clear();
        read1(n);
        rep1(i,,n) B[i] = Next[i] = ;
        rep1(i,,n) read1(a[i]);
        rep1(i,,n){
            if(a[i] >= a[i-] && a[i-] >= a[i-]){
                int x = mp[MK(MK(a[i],a[i-]),a[i-])];
                if(x == ) update(i,);
                else Next[x] = i;//递推
                mp[MK(MK(a[i],a[i-]),a[i-])] = i;
            }
        }
        read1(m);
        rep0(i,,m) read2(x[i],y[i]),o[i] = i;
        sort(o,o+m,cmp);
        int p = ;
        rep0(i,,m){
            int l = x[o[i]],r = y[o[i]];
            while(p <= l+){
                if(Next[p]) update(Next[p],);//递推到下一个~~
                p++;
            }
            if(r > l + ) ans[o[i]] = query(r) - query(l+);
            else ans[o[i]] = ;
        }
        rep0(i,,m){
            printf("%d\n",ans[i]);
        }
    }
    return ;
}