12. 17 哈理工网络赛

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 line consists of one positive 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 number which 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

题目分析 ： 类似于LIS，也可以采取维护一个最大的和的序列，但是有一点不同的是，就是在后续插入一个值后，要讲该值后面的元素中键值大于插入的，但是和却小于插入的和的 全部删除掉， 这里用 map 来维护 。

代码示例 ：

#include <cstdio>
#include <iostream>
#include <map>
using namespace std;
const int eps = 1e5+5;

int a[eps], b[eps];
map<int, int>mp;

int main (){
	int t;
	int n;
	
	cin >> t;
	while(t--){
		cin >> n;
		for(int i = 1; i <= n; i++){
			scanf("%d", &a[i]);
		}
		for(int i = 1; i <= n; i++){
			scanf("%d", &b[i]);
		}
		mp.clear();
		mp[0] = 0;
		map<int, int>::iterator it;
		int ans = -1;
		for(int i = 1; i <= n; i++){
			for(it = mp.begin(); it != mp.end(); it++){
				if (a[i] < it->first) break;
			}
			it--;
			int p = it->second + b[i];
			if (mp.count(a[i])) {
				mp[a[i]] = max(mp[a[i]], p);
			}
			else mp[a[i]] = p;
			ans = max(ans, mp[a[i]]);
			
			for(it = mp.begin(); it != mp.end(); it++){
				if (it->first > a[i] && it->second <= mp[a[i]]) {
					mp.erase(it);
					it--;   // 坑啊，删除一个元素后，it 所指向的是删除元素的下个位置，这样以来的话，出这个循环it在加加，就会造成无线循环的局面
				}

			}
		}
//		for(it = mp.begin(); it != mp.end(); it++){
//			printf("%d  %d\n", it->first, it->second);
//		}
		printf("%d\n", ans);
	}
	
	
	return 0;
}
/*
10
7
3 7 2 9 6 8 5
1 1 2 1 1 2 2
*/