zoukankan      html  css  js  c++  java
  • HDU 3333 Turing Tree(离线树状数组)

    Turing Tree

    Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 5014    Accepted Submission(s): 1777

    Problem Description
    After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again...

    Now given a sequence of N numbers A1, A2, ..., AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, ..., Aj.
     
    Input
    The first line is an integer T (1 ≤ T ≤ 10), indecating the number of testcases below.
    For each case, the input format will be like this:
    * Line 1: N (1 ≤ N ≤ 30,000).
    * Line 2: N integers A1, A2, ..., AN (0 ≤ Ai ≤ 1,000,000,000).
    * Line 3: Q (1 ≤ Q ≤ 100,000), the number of Queries.
    * Next Q lines: each line contains 2 integers i, j representing a Query (1 ≤ i ≤ j ≤ N).
     
    Output
    For each Query, print the sum of distinct values of the specified subsequence in one line.
     
    Sample Input
    2
     
     
    3
    1 1 4
    2
    1 2
    2 3
    5
    1 1 2 1 3
    3
    1 5
    2 4
    3 5
     
    Sample Output
    1
    5
    6
    3
    6

    题目链接:HDU 3333

    对于莫队式暴力就没啥好说的了,时间慢不说代码量也比较大,还是离线+BIT快

    代码:

    #include <stdio.h>
    #include <bits/stdc++.h>
    using namespace std;
    #define INF 0x3f3f3f3f
    #define LC(x) (x<<1)
    #define RC(x) ((x<<1)+1)
    #define MID(x,y) ((x+y)>>1)
    #define CLR(arr,val) memset(arr,val,sizeof(arr))
    #define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
    typedef pair<int, int> pii;
    typedef long long LL;
    const double PI = acos(-1.0);
    const int N = 30010;
    const int M = 200010;
    struct info
    {
        int k, l, r, flag, id;
        info() {}
        info(int _k, int _l, int _r, int _flag, int _id): k(_k), l(_l), r(_r), flag(_flag), id(_id) { }
        bool operator<(const info &rhs)const
        {
            return k < rhs.k;
        }
    };
    info Q[M];
    LL T[N], ans[M >> 1];
    int arr[N];
    map<int, int>last;
    
    void init()
    {
        CLR(T, 0);
        CLR(ans, 0);
        last.clear();
    }
    void add(int k, LL v)
    {
        while (k < N)
        {
            T[k] += v;
            k += (k & -k);
        }
    }
    LL getsum(int k)
    {
        LL ret = 0;
        while (k)
        {
            ret += T[k];
            k -= (k & -k);
        }
        return ret;
    }
    int main(void)
    {
        int tcase, n, m, i;
        scanf("%d", &tcase);
        while (tcase--)
        {
            init();
            scanf("%d", &n);
            for (i = 1; i <= n; ++i)
                scanf("%d", &arr[i]);
            scanf("%d", &m);
            int qcnt = 0;
            for (i = 0; i < m; ++i)
            {
                int l, r;
                scanf("%d%d", &l, &r);
                Q[qcnt++] = info(l, l, r, 0, i);
                Q[qcnt++] = info(r, l, r, 1, i);
            }
            sort(Q, Q + qcnt);
            int x = 1;
            for (i = 0; i < qcnt; ++i)
            {
                while (x <= Q[i].k)
                {
                    if (last[arr[x]])
                        add(last[arr[x]], -arr[x]);
                    add(x, arr[x]);
                    last[arr[x]] = x;
                    ++x;
                }
                if (Q[i].flag)
                    ans[Q[i].id] += getsum(Q[i].r) - getsum(Q[i].l - 1);
                else
                {
                    add(x - 1, -arr[x - 1]);
                    ans[Q[i].id] -= getsum(Q[i].r) - getsum(Q[i].l - 1);
                    add(x - 1, arr[x - 1]);
                }
            }
            for (i = 0; i < m; ++i)
                printf("%I64d
    ", ans[i]);
        }
        return 0;
    }
  • 相关阅读:
    一道亲戚的生物学改题
    【水】强化16题解
    【我为标程写注释】最大值最小化
    【我为标程写注释】卢斯进制
    oracle 解锁表
    Oracle存储过程根据指定日期返回(N个)工作日的时间
    NPOI_2.1.3_学习记录(6)-Excel中设置小数、百分比、货币、日期、科学计数法和金额大写
    NPOI_2.1.3_学习记录(5)-创建Excel的页眉页脚
    NPOI_2.1.3_学习记录(4)-Excel中单元格的复制
    NPOI_2.1.3_学习记录(2)-在Excel中创建工作表(Sheet)
  • 原文地址:https://www.cnblogs.com/Blackops/p/6390957.html
Copyright © 2011-2022 走看看