题意：给你正视和侧视图，求最多多少个，最少多少个

思路：贪心的思想，求最少的时候：因为可以想象着移动，尽量让两个视图的重叠，所以我们统计每个视图不同高度的个数，然后计算，至于的话，就是每次拿正视图的高度去匹配侧视求最大

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 1000;

int k;
int view[2][MAXN];

int main() {
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &k);
		memset(view, 0, sizeof(view));
		for (int i = 0; i < 2; i++)
			for (int j = 0; j < k; j++) {
				int x;
				scanf("%d", &x);
				view[i][x]++;
			}
		int Min = 0, Max = 0;
		for (int i = 1; i < MAXN; i++) 
			Min += i * max(view[0][i], view[1][i]);
		for (int i = 1; i < MAXN; i++)
			for (int j = 1; j < MAXN; j++) 
				Max += min(i, j)*view[0][i]*view[1][j];
		printf("Matty needs at least %d blocks, and can add at most %d extra blocks.\n", Min, Max-Min);
	}
	return 0;
}