UVA - 434 Matty's Blocks

题意：给你正视和側视图，求最多多少个，最少多少个

思路：贪心的思想。求最少的时候：由于能够想象着移动，尽量让两个视图的重叠。所以我们统计每一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側视求最大

#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;
}