测评会员优惠活动进行中 · 开通 VIP,有效期内测评不限次 VIP 优惠中 · 测评不限次 立即查看

A14962. Best Pair

编程题 普及/提高-

题目描述

You are given an array $a$ of length $n$ . Let $cnt_x$ be the number of elements from the array which are equal to $x$ . Let's also define $f(x, y)$ as $(cnt_x + cnt_y) \cdot (x + y)$ .

Also you are given $m$ bad pairs $(x_i, y_i)$ . Note that if $(x, y)$ is a bad pair, then $(y, x)$ is also bad.

Your task is to find the maximum value of $f(u, v)$ over all pairs $(u, v)$ , such that $u \neq v$ , that this pair is not bad, and also that $u$ and $v$ each occur in the array $a$ . It is guaranteed that such a pair exists.

输入格式

The first line contains a single integer $t$ ( $1 \le t \le 10\,000$ ) — the number of test cases.

The first line of each test case contains two integers $n$ and $m$ ( $2 \le n \le 3 \cdot 10^5$ , $0 \le m \le 3 \cdot 10^5$ ) — the length of the array and the number of bad pairs.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ( $1 \le a_i \le 10^9$ ) — elements of the array.

The $i$ -th of the next $m$ lines contains two integers $x_i$ and $y_i$ ( $1 \le x_i < y_i \le 10^9$ ), which represent a bad pair. It is guaranteed that no bad pair occurs twice in the input. It is also guaranteed that $cnt_{x_i} > 0$ and $cnt_{y_i} > 0$ .

It is guaranteed that for each test case there is a pair of integers $(u, v)$ , $u \ne v$ , that is not bad, and such that both of these numbers occur in $a$ .

It is guaranteed that the total sum of $n$ and the total sum of $m$ don't exceed $3 \cdot 10^5$ .

输出格式

For each test case print a single integer — the answer to the problem.

输入输出样例

输入 #1
3
6 1
6 3 6 7 3 3
3 6
2 0
3 4
7 4
1 2 2 3 1 5 1
1 5
3 5
1 3
2 5
输出 #1
40
14
15

说明/提示

In the first test case $3$ , $6$ , $7$ occur in the array.

- $f(3, 6) = (cnt_3 + cnt_6) \cdot (3 + 6) = (3 + 2) \cdot (3 + 6) = 45$ . But $(3, 6)$ is bad so we ignore it.
- $f(3, 7) = (cnt_3 + cnt_7) \cdot (3 + 7) = (3 + 1) \cdot (3 + 7) = 40$ .
- $f(6, 7) = (cnt_6 + cnt_7) \cdot (6 + 7) = (2 + 1) \cdot (6 + 7) = 39$ .

The answer to the problem is $\max(40, 39) = 40$ .
上一题 去做题 下一题