标签:can div 优秀 turn blog tor struct bool ++
顶点一定在凸包上,我们枚举对角线,观察到固定一个点后,随着另一个点的增加,剩下两个点的最优位置一定是单调的,于是就得到了一个优秀的O(n^2)做法。
#include <cstdio> #include <algorithm> const int N = 2005; int n,r,p1,p2,q[N]; double ans; struct nd { double x,y; bool operator < (const nd &b) const {return x == b.x ? y < b.y : x < b.x;} nd operator - (const nd &b) const {return (nd){x-b.x,y-b.y};} double operator * (const nd &b) const {return x*b.y-y*b.x;} }a[N]; double cj(int x, int y, int z) {return (a[x]-a[y])*(a[z]-a[y]);} int main() { scanf("%d", &n); for(int i = 1; i <= n; i++) scanf("%lf%lf", &a[i].x, &a[i].y); std::sort(a+1, a+1+n); for(int i = 1; i <= n; i++) { while(r > 1 && cj(q[r], q[r-1], i) >= 0) r--; q[++r] = i; } int rr = r; for(int i = n-1; i >= 1; i--) { while(r > rr && cj(q[r], q[r-1], i) >= 0) r--; q[++r] = i; } for(int i = 1; i < r; i++) printf("%.0f %.0f\n", a[q[i]].x, a[q[i]].y); for(int i = 1; i < r; i++) { int p1 = i+1, p2 = i+3; for(int j = i+2; j < r-1; j++) { while(p1 < j-1 && cj(q[j],q[i],q[p1]) <= cj(q[j],q[i],q[p1+1])) p1++; while(p2 <= j || (p2 < r-1 && cj(q[p2],q[i],q[j]) <= cj(q[p2+1],q[i],q[j]))) p2++; ans = std::max(ans, cj(q[j],q[i],q[p1])+cj(q[p2],q[i],q[j])); } } printf("%.3f", ans/2); return 0; }
标签:can div 优秀 turn blog tor struct bool ++
原文地址:http://www.cnblogs.com/juruolty/p/6216118.html