CF2027A-Rectangle Arrangement

CF2027A-Rectangle Arrangement

题目大意

一个无限大的平面上，给你 \(n\) 个矩形，你可以在平面上任意放置矩形，要求最小矩形覆盖周长。

题解

考虑贪心做法，将所有矩形左下角都放在 \((0,0)\) 上。记录最长的 \(x\) 坐标和 \(y\) 坐标。此时面积就是 \(2*(l+r)\) 。

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ls(p) (p << 1)
#define rs(p) (ls(p) ^ 1)
typedef pair<int, int> pii;
const int INF = 1e16;
const int N = 3e5 + 5;

inline int read();
inline void solve()
{
    int n = read(), l = 0, r = 0;
    for (int i = 1; i <= n; ++i)
    {
        int x = read(), y = read();
        l = max(l, x);
        r = max(r, y);
    }
    printf("%lld\n", 2 * (l + r));
}

signed main()
{
    int T = read();
    while (T--)
        solve();
    return 0;
}

inline int read()
{
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = (x << 1) + (x << 3) + (ch ^ 48);
        ch = getchar();
    }
    return x * f;
}