HDU 5654 (树状数组 离散化)

xiaoxin and his watermelon candy

Time Limit: 4000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 243    Accepted Submission(s): 64

Problem Description

During his six grade summer vacation, xiaoxin got lots of watermelon candies from his leader when he did his internship at Tencent. Each watermelon candy has it's sweetness which denoted by an integer number.

xiaoxin is very smart since he was a child. He arrange these candies in a line and at each time before eating candies, he selects three continuous watermelon candies from a specific range [L, R] to eat and the chosen triplet must satisfies:

if he chooses a triplet  (ai,aj,ak)  then:
1.  j=i+1,k=j+1
2.   ai≤aj≤ak

Your task is to calculate how many different ways xiaoxin can choose a triplet in range [L, R]?
two triplets  (a0,a1,a2)  and  (b0,b1,b2)  are thought as different if and only if:
a0≠b0  or  a1≠b1  or  a2≠b2

 

Input

This problem has multi test cases. First line contains a single integer  T(T≤10)  which represents the number of test cases.

For each test case, the first line contains a single integer  n(1≤n≤200,000) which represents number of watermelon candies and the following line contains  n integer numbers which are given in the order same with xiaoxin arranged them from left to right.
The third line is an integer  Q(1≤200,000)  which is the number of queries. In the following  Q  lines, each line contains two space seperated integers  l,r(1≤l≤r≤n)  which represents the range [l, r].

 

Output

For each query, print an integer which represents the number of ways xiaoxin can choose a triplet.

 

Sample Input

   
   
   
   

    
    
    
    1
5
1 2 3 4 5
3
1 3
1 4
1 5
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
3
   
   
   
   

 

题意:

求某个区间内所有的连续递增三元组的个数,只要三元组中有一个数字不同就被认为是不同的三元组.

首先把数字离散化,然后对于某一个三元组,假设三个数分别是a,b,c,如果abc满足不严格的递增,

可以设这个三元组等于a*1e12+b*1e6+c,然后继续对三元组对应的数组离散化,然后只需要离线处理询问

的区间,拍完序之后用树状数组或者线段树统计区间只出现一次的数字的个数.

#include <bits/stdc++.h>
using namespace std;
#define maxn 211111
const long long num1 = 1e12;
const long long num2 = 1e6;

long long a[maxn];
int last [maxn], ans[maxn];
struct node {
    long long num;
    int pos;
    bool operator < (const node &a) const {
        return num < a.num;
    }
}b[maxn];
int n, q;
struct query {
    int l, r, id;
    bool operator < (const query &a) const {
        return l > a.l;
    }
}qu[maxn];

int c[maxn];
int lowbit (int x) {
    return x&(-x);
}

void add (int pos, int num) {
    for (int i = pos; i <= n-2; i += lowbit (i)) {
        c[i] += num;
    }
}

int sum (int pos) {
    int ans = 0;
    for (int i = pos; i > 0; i -= lowbit(i)) {
        ans += c[i];
    }
    return ans;
}

int main () {
    //freopen ("in.txt", "r", stdin);
    int t;
    scanf ("%d", &t);
    while (t--) {
        scanf ("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf ("%lld", &a[i]);
            b[i].pos = i,  b[i].num = a[i];
        }
        sort (b+1, b+1+n);
        scanf ("%d", &q);
        for (int i = 1; i <= q; i++) {
            scanf ("%d%d", &qu[i].l, &qu[i].r);
            qu[i].r -= 2;
            qu[i].id = i;
        }
        if (n == 1 || n == 2) {
            for (int i = 1; i <= q; i++) {
                printf ("0\n");
            }
            continue;
        }
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            if (i > 1 && b[i].num == b[i-1].num) {
                a[b[i].pos] = cnt;
            }
            else
                a[b[i].pos] = ++cnt;
        }
        for (int i = 1; i <= n-2; i++) { 
            if (a[i]<=a[i+1] && a[i+1]<=a[i+2])
                b[i].num = a[i]*num1+a[i+1]*num2+a[i+2];
            else
                b[i].num = 0;
            b[i].pos = i;
        }
        sort (b+1, b+1+n-2);
        cnt = 0;
        for (int i = 1; i <= n-2; i++) {
            if (b[i].num == 0) {
                a[b[i].pos] = 0;
            }
            else if (i > 1 && b[i].num == b[i-1].num) {
                a[b[i].pos] = cnt;
            }
            else
                a[b[i].pos] = ++cnt;
        } 
        sort (qu+1, qu+1+q);
        memset (c, 0, sizeof c);
        memset (last, -1, sizeof last);
        for (int i = n-2; i >= qu[1].l; i--) {
            if (a[i] == 0)
                continue;
            if (last[a[i]] == -1) {
                last[a[i]] = i;
                add (i, 1);
            }
            else {
                add (last[a[i]], -1);
                add (i, 1);
                last[a[i]] = i;
            }
        } 
        for (int i = 1; i <= q; i++) {
            if (i > 1 && qu[i].l < qu[i-1].l) {
                for (int j = qu[i-1].l-1; j >= qu[i].l; j--) {
                    if (a[j] == 0)
                        continue;
                    else if (last[a[j]] == -1) {
                        last[a[j]] = j;
                        add (j, 1);
                    }
                    else {
                        add (last[a[j]], -1);
                        add (j, 1);
                        last[a[j]] = j;
                    }
                }
            }
            ans[qu[i].id] = sum (qu[i].r);
        }
        for (int i = 1; i <= q; i++)
            printf ("%d\n", ans[i]);
    }
    return 0;
}