poj 3714 Raid

http://poj.org/problem?id=3714

求平面中给定的2*N个点, 距离最短的两点距离(两个点必须分别属于两个集合)

套用了别人的求平面最短两点距离的模板, 囧

//model start#include<stdlib.h>#include<stdio.h>#include<string.h>#include<math.h>const int N = 200002;const double MAX = 10e100, eps = 0.00001; //mark!struct Point {double x, y;int index;short set;}a[N], b[N], c[N];inline double min(double p, double q) {return p < q ? p : q;}inline double dis(Point p, Point q) {double x1 = p.x - q.x;double y1 = p.y - q.y; return sqrt(x1*x1 + y1*y1);}int cmp_x(const void *p, const void *q) {double temp = ((Point *)p)->x - ((Point *)q)->x;if(temp>0) {return 1; } else if(fabs(temp) < eps) {return 0;} else { return -1;}}int cmp_y(const void *p, const void *q) {double temp = ((Point *)p)->y - ((Point *)q)->y;if(temp > 0) {return 1; } else if(fabs(temp)<eps){return 0;} else {return -1;}}void merge(Point p[], Point q[], int s, int m, int t) {int i, j, k;for(i=s,j=m+1,k=s; i<=m&&j<=t;) {if(q[i].y > q[j].y) {p[k++] = q[j];++j;} else {p[k++] = q[i];++i;}}while(i <= m) {p[k++] = q[i];++i;} while(j <= t) {p[k++] = q[j];++j;}}double closest(Point *a, Point *b, Point *c, int p, int q) {if(q-p == 1) {if(a[p].set != a[q].set)return dis(a[p], a[q]);elsereturn MAX;}if(q-p == 2) {double x1, x2, x3;if(a[p].set != a[q].set) x1 = dis(a[p], a[q]);elsex1 = MAX;if(a[p+1].set != a[q].set) x2 = dis(a[p+1], a[q]);elsex2 = MAX;if(a[p].set != a[p+1].set) x3 = dis(a[p], a[p+1]);elsex3 = MAX;double m = min(x1, x2);return min(m, x3);}int i, j, k, m = (p+q)>>1;double d1, d2;for(i=p,j=p,k=m+1; i<=q; ++i) {if(b[i].index <= m) {c[j++] = b[i];} else {c[k++] = b[i];}}d1 = closest(a, c, b, p, m);d2 = closest(a, c, b, m+1, q);double dm = min(d1, d2);merge(b, c, p, m, q);for(i=p,k=p; i<=q; ++i) {if(fabs(b[i].x - b[m].x) < dm) {c[k++] = b[i];}}for(i=p; i<k; ++i) {for(j=i+1; j<k&&c[j].y-c[i].y<dm; ++j) {if(c[j].set != c[i].set) {double temp = dis(c[j], c[i]);if(temp < dm) {dm = temp;}}}}return dm;} //model endint main() {//freopen("in.txt", "r", stdin);int n, i, t;double ans;scanf("%d",&t);while(t--) {scanf("%d", &n);int mid = n;n = n<<1;for(i=0; i<n; ++i) {scanf("%lf%lf", &a[i].x, &a[i].y);if(i < mid)a[i].set = 0;elsea[i].set = 1;}qsort(a, n, sizeof(a[0]), cmp_x);for(i=0; i<n; ++i) {a[i].index = i;}memcpy(b, a, sizeof(a[0])*n);qsort(b, n, sizeof(b[0]), cmp_y);ans = closest(a, b, c, 0, n-1);printf("%.3lf/n", ans);}return 0;}

 

求平面中N个点的最短距离的模板

//求平面中N个点, 距离最近的两点的距离//model start#include<stdlib.h>#include<stdio.h>#include<string.h>#include<math.h>const int N = 100005;const double MAX = 10e100, eps = 0.00001; //mark!struct Point {double x, y;int index;}a[N], b[N], c[N];inline double min(double p, double q) {return p < q ? p : q;}inline double dis(Point p, Point q) {double x1 = p.x - q.x;double y1 = p.y - q.y; return sqrt(x1*x1 + y1*y1);}int cmp_x(const void *p, const void *q){ double temp = ((Point *)p)->x - ((Point *)q)->x;if(temp>0) {return 1; } else if(fabs(temp) < eps) {return 0;} else { return -1;}}int cmp_y(const void *p, const void *q) {double temp = ((Point *)p)->y - ((Point *)q)->y;if(temp > 0) {return 1; } else if(fabs(temp)<eps){return 0;} else {return -1;}}void merge(Point p[], Point q[], int s, int m, int t) {int i, j, k;for(i=s,j=m+1,k=s; i<=m&&j<=t;) {if(q[i].y > q[j].y) {p[k++] = q[j];++j;} else {p[k++] = q[i];++i;}}while(i <= m) {p[k++] = q[i];++i;} while(j <= t) {p[k++] = q[j];++j;}}double closest(Point *a, Point *b, Point *c, int p, int q) {if(q-p == 1) {return dis(a[p], a[q]);}if(q-p == 2) {double x1 = dis(a[p], a[q]);double x2 = dis(a[p+1], a[q]);double x3 = dis(a[p], a[p+1]);double m = min(x1, x2);return min(m, x3);}int i, j, k, m = (p+q)>>1;double d1, d2;for(i=p,j=p,k=m+1; i<=q; ++i) {if(b[i].index <= m) {c[j++] = b[i];} else {c[k++] = b[i];}}d1 = closest(a, c, b, p, m);d2 = closest(a, c, b, m+1, q);double dm = min(d1, d2);merge(b, c, p, m, q);for(i=p,k=p; i<=q; ++i) {if(fabs(b[i].x - b[m].x) < dm) {c[k++] = b[i];}}for(i=p; i<k; ++i) {for(j=i+1; j<k&&c[j].y-c[i].y<dm; ++j) {double temp = dis(c[j], c[i]);if(temp < dm) {dm = temp;}}}return dm;} //model endint main() {//freopen("in.txt", "r", stdin);int n, i;double ans;while(scanf("%d", &n) != EOF) {for(i=0; i<n; ++i) {scanf("%lf%lf", &a[i].x, &a[i].y);}qsort(a, n, sizeof(a[0]), cmp_x);for(i=0; i<n; ++i) {a[i].index = i;}memcpy(b, a, sizeof(a[0])*n);qsort(b, n, sizeof(b[0]), cmp_y);ans = closest(a, b, c, 0, n-1);//ans为距离的最小值printf("%.3lf/n", ans);}return 0;}