题意：就是给N个点的坐标，然后求任意两个点距离的平方最大的值
枚举超时。
当明白了 最远距离的两个点一定在凸包上，一切就好办了。求出凸包，然后枚举
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> using namespace std; const double eps = 1e-6; const int Max = 50000 + 10; struct Node { int x, y; }; Node node[Max], ch[Max]; int cmp(Node tempx, Node tempy) { if(tempx.y == tempy.y) return tempx.x < tempy.x; return tempx.y < tempy.y; } int xmult(Node p1, Node p2, Node p3) { return (p2.x - p1.x) * (p3.y - p1.y) - (p3.x - p1.x) * (p2.y - p1.y); } int andrew(int n) { int len, top = 1; ch[0] = node[0]; ch[1] = node[1]; for(int i = 2; i < n; i++) { while(top && xmult(ch[top - 1], ch[top], node[i]) <= 0) top--; ch[ ++top ] = node[i]; } len = top; ch[ ++top ] = node[n - 2]; for(int i = n - 3; i >= 0; i--) { while(len != top && xmult(ch[top - 1], ch[top], node[i]) <= 0) top--; ch[ ++top ] = node[i]; } return top; } int dist(Node p1, Node p2) { return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y) ; } int get_max(int x, int y) { if(x - y > 0) return x; return y; } int main() { int n; while( scanf("%d", &n) != EOF) { for(int i = 0; i < n; i++) scanf("%d%d", &node[i].x, &node[i].y); sort(node, node + n, cmp); int top = andrew(n); int ans = 0; if(top > 2) { for(int i = 0; i < top; i++) { for(int j = 0; j < top; j++) { ans = get_max(ans, dist(ch[i], ch[j])); } } } else { ans = dist(ch[0], ch[1]); } printf("%d ", ans); } return 0; }