Codeforces 460E Roland and Rose(暴力)

题目链接：Codeforces 460E Roland and Rose

题目大意：在以原点为圆心，半径为R的局域内选择N个整数点，使得N个点中两两距离的平方和最大。

解题思路：R最大为30。那么事实上距离圆心距离最大的整数点只是12个最多，直接暴力枚举。

#include 
#include 
#include 
#include 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 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
", ans);for (int i = 0; i < N; i++)printf("%d %d
", vec[rec[i]].x, vec[rec[i]].y);return 0;
}