hdu 1007 Quoit Design (最近点对、分治)

hdu 1007 Quoit Design (最近点对)

http://acm.hdu.edu.cn/showproblem.php?pid=1007

“最近点对”的经典入门题

所用到的思想和“归并排序”类似， 分治再合并。

 

<pre class="cpp" name="code">#include "stdio.h" #include "cmath" #include "iostream" #include "algorithm" using namespace std; struct point {double x, y;}; point p[100003]; point tp[100003]; bool cmp_x(const point & a, const point & b) { return a.x < b.x; } bool cmp_y(const point & a, const point & b) { return a.y < b.y; } double min(double a, double b) { return a < b ? a : b; } double dis(const point & a, const point & b) { return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)); } double solve(int lf, int rt, int n) //left, right { if(lf == rt) return 1e9; //只有一个点时，距离设为无穷大 if(rt - lf == 1) return dis(p[lf], p[rt]); int mid = (lf + rt) >> 1; double ans = min(solve(lf, mid, n), solve(mid + 1, rt, n)); int lp = 0, i, j; for(i = mid; i >= lf && p[i].x - p[mid].x < ans; i--) { tp[lp].x = p[i].x; tp[lp++].y = p[i].y; } for(i = mid + 1; i <= rt && p[i].x - p[mid].x < ans; i++) { tp[lp].x = p[i].x; tp[lp++].y = p[i].y; } sort(tp, tp + lp, cmp_y); for(i = 0; i < lp; i++) for(j = i + 1; j - i < 7 && j < lp; j++) ans = min(ans, dis(tp[i], tp[j])); return ans; } int main() { int n, i; while(scanf("%d", &n) && n) { for(i = 1; i <= n; i++) scanf("%lf%lf", &p[i].x, &p[i].y); sort(p + 1, p + n + 1, cmp_x); printf("%.2lf\n", solve(1, n, n) / 2.0); } return 0; }