pku 2187(最远点对)

最新推荐文章于 2019-01-08 01:48:28 发布

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最新推荐文章于 2019-01-08 01:48:28 发布

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原文链接：http://www.cnblogs.com/try86/archive/2012/04/27/2472927.html

本文介绍了一种利用凸包算法结合枚举的方法来寻找一组二维平面上的点中距离最远的两个点。该算法首先通过Graham扫描法找出构成凸包的点集，然后遍历这些点来找到距离最远的点对。

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 1 /*
 2 *  题目要求：求一组点中距离最远的一对的距离 
 3 *  解法：凸包+枚举 
 4 */
 5 
 6 #include <cmath>
 7 #include <cstdio>
 8 #include <cstdlib>
 9 #include <iostream>
10 
11 using namespace std;
12 
13 const int N = 50005;
14 
15 struct point {
16     int x;
17     int y;
18 }p[N], stack[N];
19 
20 int dis(point A, point B) {
21     return (A.x-B.x)*(A.x-B.x) + (A.y-B.y)*(A.y-B.y);
22 }
23 
24 int crossProd(point A, point B, point C) {
25     return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
26 }
27 
28 int cmp(const void *a, const void *b) {
29     point *c = (point *)a;
30     point *d = (point *)b;
31     int k = crossProd(p[0], *c, *d);
32     if (k<0 || (!k&&dis(p[0], *c)>dis(p[0], *d))) return 1;
33     return -1;
34 }
35 
36 int Graham(int n) {
37     int x = p[0].x;
38     int y = p[0].y;
39     int mi = 0;
40     for (int i=1; i<n; ++i) {
41         if (p[i].y<y || (p[i].y==y&&p[i].x<x)) {
42             x = p[i].x;
43             y = p[i].y;
44             mi = i;
45         }
46     } 
47     point tmp = p[mi];
48     p[mi] = p[0];
49     p[0] = tmp;
50     qsort(p+1, n-1, sizeof(point), cmp);
51     p[n] = p[0];
52     for (int i=0; i<2; ++i) stack[i] = p[i];
53     int top = 1;
54     for (int i=2; i<n; ++i) {
55         while (crossProd(stack[top-1], stack[top], p[i])<=0 && top>=1) --top;
56         stack[++top] = p[i];
57     }
58     return top;
59 }
60 
61 int solve(int n) {
62     int top = Graham(n);
63     int s, maxL = 0;
64     for (int i=0; i<top; ++i) {
65         for (int j=i+1; j<=top; ++j) {
66             s = dis(stack[i], stack[j]);
67             if (s > maxL) maxL = s;
68         }
69     }
70     return maxL;
71 }
72 
73 int  main() {
74     int n;
75     while (scanf("%d", &n) != EOF) {
76         for (int i=0; i<n; ++i) scanf ("%d%d", &p[i].x, &p[i].y);
77         int ans = solve(n);
78         printf ("%d\n", ans);
79     }
80     return 0;
81 }