zoukankan      html  css  js  c++  java
  • hdu 3333 Turing Tree(树状数组 + 离线查询)

    Turing Tree

    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

    为了写一道cf的题专门来做了这题,题意是求区间内不同的数之和,方法是离线查询:将要询问的区间保存下来,按照区间右端点的大小排序,这样的话在查询后一个区间时,前面已经处理完了(将已出现过的数字置为0,只保留最后一个),用树状数组简单的区间求和。

    #include <bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int maxn = 1e5 + 10;
    ll a[maxn];
    ll c[maxn];
    int n;
    
    struct node {
        int l, r, id;
        ll ans = 0;
    }p[maxn];
    
    inline ll lowbit(ll x) {
        return x & (-x);
    }
    
    ll sum(int x) {
        ll ret = 0;
        while(x > 0) {
            ret += c[x]; x -= lowbit(x);
        }
        return ret;
    }
    
    void add(int x, ll d) {
        while(x <= n) {
            c[x] += d; x += lowbit(x);
        }
    }
    
    bool cmp1(node a, node b) {
        return a.r < b.r;
    }
    
    bool cmp2(node a, node b) {
        return a.id < b.id;
    }
    
    int main() {
        int t;
        scanf("%d", &t);
        while(t--) {
            scanf("%d", &n);
            for(int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
            int m;
            scanf("%d", &m);
            for(int i = 1; i <= m; i++) {
                scanf("%d %d", &p[i].l, &p[i].r);
                p[i].id = i;
            }
            sort(p + 1, p + 1 + m, cmp1);
            memset(c, 0, sizeof(c));
            map<int, int> mp;
            int pos = 1;
            for(int i = 1; i <= n; i++) {
                if(mp[a[i]]) {
                    add(mp[a[i]], -a[i]);
                    a[mp[a[i]]] = 0;
                }
                mp[a[i]] = i;
                add(i, a[i]);
                for( ; pos <= m && p[pos].r == i; pos++) {
                    p[pos].ans = sum(p[pos].r) - sum(p[pos].l - 1);
                }
            }
            sort(p + 1, p + 1 + m, cmp2);
            for(int i = 1; i <= m; i++) {
                printf("%I64d
    ", p[i].ans);
            }
        }
    }
  • 相关阅读:
    字符串去重
    你必须懂的 T4 模板:深入浅出
    解决T4模板的程序集引用的五种方案
    table 合并行和列
    porwedesigner 去掉引号
    面向对象JS基础
    19套最新的免费图标字体集
    推荐 15 款很棒的文本编辑器
    13个JavaScript图表(JS图表)图形绘制插件
    推荐10款免费而优秀的图表插件
  • 原文地址:https://www.cnblogs.com/lonewanderer/p/5767333.html
Copyright © 2011-2022 走看看