Codeforces 460E Roland and Rose (violent), codeforces460e

Codeforces 460E Roland and Rose (violent), codeforces460e

Link: Codeforces 460E Roland and Rose

N Integer Points are selected in the local area where the origin is the center and the radius is R, so that the sum of squares of the distance between the two points is the largest.

Solution: The maximum R value is 30. In fact, there are only 12 Integer Points with the largest distance from the center of the circle, and the enumeration is directly violent.

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;struct point {    int x, y;    point (int x = 0, int y = 0) {        this->x = x;        this->y = y;    }};int N, R, M, ans, pos[10], rec[10];vector<point> vec;inline int dis (int x, int y) {    return x * x + y * y;}inline bool cmp (const point& a, const point& b) {    return dis(a.x, a.y) > dis(b.x, b.y);}void init () {    scanf("%d%d", &N, &R);    for (int i = -R; i <= R; i++) {        for (int j = -R; j <= R; j++)  {            if (i * i + j * j <= R * R)                vec.push_back(point(i, j));        }    }    ans = 0;    M = min((int)vec.size(), 18);    sort(vec.begin(), vec.end(), cmp);}void dfs (int d, int f, int s) {    if (d == N) {        if (s > ans) {            ans = s;            memcpy(rec, pos, sizeof(pos));        }        return;    }    for (int i = f; i < M; i++) {        int add = 0;        for (int j = 0; j < d; j++)            add += dis(vec[pos[j]].x - vec[i].x, vec[pos[j]].y - vec[i].y);        pos[d] = i;        dfs(d + 1, i, s + add);    }}int main () {    init();    dfs(0, 0, 0);    printf("%d\n", ans);    for (int i = 0; i < N; i++)        printf("%d %d\n", vec[rec[i]].x, vec[rec[i]].y);    return 0;}