Codeforces 460E Roland and Rose(暴力)

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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，那么其实距离圆心距离最大的整数点不过12个最多，直接暴力枚举。

#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;}

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