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题意:n*4个点,求覆盖所有点的最小矩形面积;
思路:凸包+旋转卡壳,uva10173几乎原题,暂时没写出来,贴一下斌神的码,待补;
#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <stdlib.h> #include <time.h> using namespace std; const int MAXN=5000; const double PI=acos(-1.0); struct point { int x,y; }; point list[MAXN]; int stack[MAXN],top; int cross(point p0,point p1,point p2) //计算叉积 p0p1 X p0p2 { return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x); } double dis(point p1,point p2) //计算 p1p2的 距离 { return sqrt((double)(p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y)); } bool cmp(point p1,point p2) //极角排序函数 , 角度相同则距离小的在前面 { int tmp=cross(list[0],p1,p2); if(tmp>0) return true; else if(tmp==0&&dis(list[0],p1)<dis(list[0],p2)) return true; else return false; } void init(int n) //输入,并把 最左下方的点放在 list[0] 。并且进行极角排序 { int i,k; point p0; scanf("%d%d",&list[0].x,&list[0].y); p0.x=list[0].x; p0.y=list[0].y; k=0; for(i=1;i<n;i++) { scanf("%d%d",&list[i].x,&list[i].y); if( (p0.y>list[i].y) || ((p0.y==list[i].y)&&(p0.x>list[i].x)) ) { p0.x=list[i].x; p0.y=list[i].y; k=i; } } list[k]=list[0]; list[0]=p0; sort(list+1,list+n,cmp); } void graham(int n) { int i; if(n==1) {top=0;stack[0]=0;} if(n==2) { top=1; stack[0]=0; stack[1]=1; } if(n>2) { for(i=0;i<=1;i++) stack[i]=i; top=1; for(i=2;i<n;i++) { while(top>0&&cross(list[stack[top-1]],list[stack[top]],list[i])<=0) top--; top++; stack[top]=i; } } } int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int T; int iCase = 0; scanf("%d",&T); int n; while(T--){ iCase++; scanf("%d",&n); init(4*n); graham(4*n); double ans = 1e20; stack[top+1] = 0; for(int i = 0;i <= top;i++){ double x1 = list[stack[i]].x; double y1 = list[stack[i]].y; double x2 = list[stack[i+1]].x; double y2 = list[stack[i+1]].y; double Min1 = 1e30, Max1 = -1e30; double Min2 = 1e30, Max2 = -1e30; for(int i = 0;i <= top;i++){ double x0 = list[stack[i]].x; double y0 = list[stack[i]].y; double tmp1 = -(y2-y1)*x0 + (x2-x1)*y0; double tmp2 = -(x2-x1)*x0 - (y2-y1)*y0; Min1 = min(Min1,tmp1); Max1 = max(Max1,tmp1); Min2 = min(Min2,tmp2); Max2 = max(Max2,tmp2); } double a = y2-y1; double b = x2-x1; ans = min(ans,(Max1-Min1)*(Max2-Min2)/(a*a+b*b)); } printf("Case #%d:\n%.0lf\n",iCase,ans); } return 0; }
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原文地址:http://www.cnblogs.com/dashuzhilin/p/4541315.html