hrbust 2362 Aggie’s Tasks【树状数组】

最新推荐文章于 2018-04-28 23:13:26 发布

Head_Hard

最新推荐文章于 2018-04-28 23:13:26 发布

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分类专栏：树状数组

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Aggie面临着一系列任务选择，每个任务都有难度和预期利润。为了最大化收益同时避免浪费时间在简单任务上，Aggie采取了一种策略：一旦接受了某个难度的任务，她就不会再接受更简单的任务。本篇介绍了一个算法解决方案，利用树状数组来帮助Aggie找到最大可能的利润。

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Aggie’s Tasks

Time Limit: 1000 MS

Memory Limit: 262144 K

Total Submit: 83(34 users)

Total Accepted: 26(20 users)

Rating:

Special Judge: No

Description

Aggie is faced with a sequence of tasks, each of which with a difficulty value d_i and an expected profit p_i. For each task, Aggie must decide whether or not to complete it. As Aggie doesn’t want to waste her time on easy tasks, once she takes a task with a difficulty d_i, she won’t take any task whose difficulty value is less than or equal to d_i.

Now Aggie needs to know the largest profits that she may gain.

Input

The first lineconsists of onepositive integer t (t≤ 10), which denotes the number of test cases.

For each test case, the first line consists one positive integer n (n≤ 100000), which denotes the number of tasks. The second line consists of n positive integers, d₁, d₂, …, d_n (d_i ≤ 100000), which is the difficulty value of each task. The third line consists of n positive integers, p₁, p₂, …, p_n (p_i ≤ 100), which is the profit that each task may bring to Aggie.

Output

For each test case, output a single numberwhich is the largest profits that Aggie may gain.

Sample Input

3 4 5 1 2

1 1 1 2 2

Sample Output

看数据范围显然不能用lcs，这里用到树状数组；

具体看代码中注释；

#include<bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 7;
int c[MAX];
struct node
{
    int x, y;
} a[MAX];
int lowbit(int i)
{
    return i & (-i);
}
void add(int i, int num) ///更新区间值，将当前价值更新到树状数组中；
{
    while(i < MAX)  ///注意此处是MAX，不是n
    {
        c[i] = max(c[i], num);  /// 比较得出最佳方案后将其在树状数组中向上将更新的值传递上去；
        i += lowbit(i);
    }
}
int Sum(int i) ///得到该节点的当前最优解；
{
    int sum = -0x3f3f3f3f;
    while(i > 0)
    {
        sum = max(sum, c[i]); /// 从该节点向下遍历子节点寻找最优解；
        i -= lowbit(i);
    }
    return sum;
}
int main()
{
    int N, n;
    scanf("%d", &N);
    while(N--)
    {
        memset(c, 0, sizeof c);
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i].x);
            a[i].x++; /// 因为下面要查询前一元素的最优解，并且数据范围可以等于0，所以++以避免（0-1）的情况
        }
        for(int i = 0; i < n; i++)
            scanf("%d", &a[i].y);
        int ans = -0x3f3f3f3f;
        for(int i = 0; i < n; i++)
        {
            int maxn = Sum(a[i].x - 1); /// 查询下一子节点的最优解；
            add(a[i].x, maxn + a[i].y); ///将（刚查过得最优解+当前价值）更新到区间中；
            ans = max(ans, maxn + a[i].y);  /// 遍历寻找最优解；
        }
        printf("%d\n", ans);
    }
}