"ARC103D" Robot Arms "structure"

The meaning of problems

Given \ (n \) points, you need to find a suitable \ (m \) and \ (D_1, d_2, ..., D_m \) , so that each one starting from the origin to go to the four directions \ (d_i \) units, eventually reaching the \ ((x_t, Y_t) \) . Output \ (m \) and \ (D \) array; for \ (t = 1 \ to n \) output direction.

\ (n-\ Leq. 3 ^ 10 \) , coordinate range \ (10 ^ 9 \)

answer

If the point \ ((x_t, Y_t) \) , \ (x_t + Y_t \) of different parity that no solution

If \ (x_t + y_t \) is even, let us \ (D_1. 1 = \) , thus converted into \ (x_t + y_t \) is an odd number.

Odd how to do? Binary structure.

A : Consider \ (d (k) = { 1, 2, 4, ..., 2 ^ k} \) such sets, can go all \ (| x | + | y | \ leq 2 ^ {k -1 + 1} \) point (and an odd)

Proof: https://blog.csdn.net/qq_37555704/article/details/83217444#_14 here moved the map.

For \ (d (k) - d (k-1) \) region (the new zone), will go directly to step (above red to violet)

For the original area can be reached, we can go inside from the inside.

Two : Consider how to find solutions.

Proof: \ (D (K) \) can come to a point satisfies \ (\ min (| x | , | y |) \ leq 2 ^ k-1 \)

Most probably prove the value \ (| = 2 ^ k \ x | = 2 ^ k-1, | | y) taken to the adjusted found not preferable.

As long as we each from \ (d (k) \) went \ (d (k-1) \) can be, to \ (d (0) \) when will succeed

By drawing the analysis, each of the large absolute value can be reduced from \ (d (k) \) come \ (D (. 1-K) \) .

Then the problem is gone. There are above us \ (d (k) \) went \ (d (k-1) \) , is actually the other way around, so pay attention to the letter output negated.

#include <algorithm>
#include <cstdio>
using namespace std;

const int N = 1010;

int n, m, x[N], y[N], d[N];
bool tg[N];

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++) {
        scanf("%d%d", x + i, y + i);
        tg[(x[i] + y[i]) & 1] = 1;
    }
    if(tg[0] & tg[1]) return puts("-1") & 0;
    if(tg[0]) d[++ m] = 1;
    for(int i = 30; ~ i; i --) d[++ m] = 1 << i;
    printf("%d\n", m);
    for(int i = 1; i <= m; i ++) printf("%d%c", d[i], " \n"[i == m]);
    for(int i = 1; i <= n; i ++, putchar('\n')) {
        for(int j = 1; j <= m; j ++) {
            if(abs(x[i]) > abs(y[i])) {
                if(x[i] > 0) x[i] -= d[j], putchar('R');
                else x[i] += d[j], putchar('L');
            } else {
                if(y[i] > 0) y[i] -= d[j], putchar('U');
                else y[i] += d[j], putchar('D');
            }
        }
    }
    return 0;
}