题目：
Quoit Design |
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
Total Submission(s): 136 Accepted Submission(s): 77 |
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Problem Description Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.
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Input The input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0.
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Output For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places.
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Sample Input 20 01 121 11 13-1.5 00 00 1.50
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Sample Output 0.710.000.75
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Author CHEN, Yue
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Source ZJCPC2004
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Recommend JGShining |
题目分析：
最小点对问题。所谓的最小点对问题就是,在n个点中找到2个点间的最短距离。这种题有两种思路：
1）直接暴力。看一看数据规模,n都在100000左右了，O(n^2)的算法，不出意外,会TLE。
2）分治。这道题用的是吉林大学的模板。直接套进去就行了。
与最小点对问题对应的是最大点对问题(不知道有没有这个名词,如果没有就当是我瞎编的吧。所谓的最大点对问题,在我的定义里就是,在n个点中找到两个点之间的最大距离)。能够产生最大距离的这两个点一定在凸包上，这时候我们只要枚举凸包上的任意两个点即可。其实这时候除了盲目枚举外，还有一种更好的算法来解决这个问题——旋转卡壳算法。
代码如下：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
/**
* 求n个点中,2个点之间的最短距离。
* 1)直接暴力。肯定会TLE
* 2)使用吉林大学的模板
*/
const int N = 100005;
const double MAX = 10e100, eps = 0.00001;
struct Point {
double x, y;
int index;
};
Point a[N], b[N], c[N];
double closest(Point *, Point *, Point *, int, int);
double dis(Point, Point);
int cmp_x(const void *, const void*);
int cmp_y(const void *, const void*);
int merge(Point *, Point *, int, int, int);
inline double min(double, double);
int main(){
int n;
while(scanf("%d",&n)!=EOF,n){
int i;
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(目标地址,起始地址,n个字节)
* 作用:从起始地址拷贝n个字节到目标地址
* 头文件: 尽量把 <cstring>引入
*
*/
memcpy(b,a,n*sizeof(a[0]));
qsort(b,n,sizeof(b[0]),cmp_y);
double ans = closest(a,b,c,0,n-1);
printf("%.2lf\n",ans/2);
}
return 0;
}
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]);
if (x1 < x2 && x1 < x3){
return x1;
}
else if (x2 < x3){
return x2;
}
else{
return x3;
}
}
int i, j, k, m = (p + q) / 2;
double d1, d2;
for (i = p, j = p, k = m + 1; i <= q; i++){
if (b[i].index <= m){
c[j++] = b[i];
}
// 数组c 左半部保存划分后左部的点, 且对y 是有序的.
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);
// 数组c 左右部分分别是对y 坐标有序的, 将其合并到b.
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];
}
}
// 找出离划分基准左右不超过dm 的部分, 且仍然对y 坐标有序.
for (i = p; i < k; i++){
for (j = i + 1; j < k && c[j].y - c[i].y < dm; j++) {
double temp = dis(c[i], c[j]);
if (temp < dm){
dm = temp;
}
}
}
return dm;
}
double dis(Point p, Point q) {
double x1 = p.x - q.x, y1 = p.y - q.y;
return sqrt(x1 * x1 + y1 * y1);
}
int 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++];
}
while (j <= t){
p[k++] = q[j++];
}
memcpy(q + s, p + s, (t - s + 1) * sizeof(p[0]));
return 0;
}
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;
}
}
inline double min(double p, double q) {
return (p > q) ? (q) : (p);
}
