科林明伦杯哈尔滨理工大学第七届程序设计团队赛-Aggie’s Tasks

Aggie’s TasksTime Limit: 1000 MSMemory Limit: 262144 KTotal Submit: 5(4 users)Total Accepted: 1(1 users)Rating: Special Judge: NoDescription

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

1

5

3 4 5 1 2

1 1 1 2 2

Sample Output

4

Source

“科林明伦杯”哈尔滨理工大学第七届程序设计团队赛

题意为，有n个任务，每个任务有各自的价值和利润，必须按顺序做任务并且下一步只能做价值比以前高的任务，问能带来的最大利润。

带权上升子序列权值和的最大值，带权LIS，不过我首先想到的是用线段树维护DP过程，后来觉得好像树状数组也可以，随性来了一发。

与普通树状数组不同的是，在单点更新和区间查询时，维护过程为寻找目前状态所能得到的最大值，也就是在增加数据时，看一遍后面跟这个点有关的点的值是否小于这个数据，如果小于则换掉，在差区间和时，也是寻找前方所存在的最大值的过程。

#include <bits/stdc++.h> using namespace std;const int N = 1e5+10;int t, n, ans, mx;int d[N], p[N], tree[N], dp[N];void add(int pos, int val) {while(pos <= mx) {tree[pos] = max(tree[pos], val);pos += pos&(-pos);}}int maxx(int pos) {int res = 0;while(pos) {res = max(res, tree[pos]);pos -= pos&(-pos);}return res;}int main() {scanf("%d", &t);while(t--) {ans = mx = 0;scanf("%d", &n);memset(tree, 0, sizeof(tree));for(int i = 1; i <= n; i++) {scanf("%d", &d[i]);mx = max(mx, d[i]);}for(int i = 1; i <= n; i++) scanf("%d", &p[i]);for(int i = 1; i <= n; i++) {dp[i] = maxx(d[i]-1) + p[i];ans = max(ans, dp[i]);add(d[i], dp[i]);}printf("%d\n", ans);}}