Educational Codeforces Round 99 部分题记录

算法竞赛
文章目录
  1. 1. E

老年复健敷衍题解。

E

ref

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#include <algorithm>
#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
int T;
ll x[4], y[4], xx[4], yy[4];
ll calc() {
ll lxMax = max(xx[0], xx[2]), lxMin = min(xx[0], xx[2]), rxMin = min(xx[1], xx[3]), rxMax = max(xx[1], xx[3]);
ll lx = max(rxMin - lxMax, 0LL), rx = max(rxMax - lxMin, 0LL);
ll lyMax = max(yy[2], yy[3]), lyMin = min(yy[2], yy[3]), ryMin = min(yy[0], yy[1]), ryMax = max(yy[0], yy[1]);
ll ly = max(ryMin - lyMax, 0LL), ry = max(ryMax - lyMin, 0LL);
ll valX = rx != 0 ? lxMax - lxMin + rxMax - rxMin : lxMin - rxMin + lxMax - rxMax;
ll valY = ry != 0 ? lyMax - lyMin + ryMax - ryMin : lyMin - ryMin + lyMax - ryMax;
if (ry >= lx && ly <= rx) return valX + valY;
else if (ly > rx) return valX + valY + 2 * (ly - rx);
else return valX + valY + 2 * (lx - ry);
}
int main() {
scanf("%d", &T);
while (T--) {
for (int i = 0; i < 4; i++) {
scanf("%lld %lld", &x[i], &y[i]);
}
int num[] = {0, 1, 2, 3};
ll ans = 0x3f3f3f3f3f3f3f3fLL;
do {
for (int i = 0; i < 4; i++) {
xx[i] = x[num[i]];
yy[i] = y[num[i]];
}
ans = min(ans, calc());
} while (next_permutation(num, num+4));
printf("%lld\n", ans);
}
return 0;
}