这题居然是WF的题， 应属于签到题。。
求一个多边形是否能被一个宽为d的矩形框住。
可以求一下凸包，然后枚举每条凸包的边，找出距离最远的点。
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 100000 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 const int MAXN=1550; 18 struct point 19 { 20 double x,y; 21 point(double x=0,double y=0):x(x),y(y){} 22 }p[N],ch[N]; 23 typedef point pointt; 24 point operator -(point a,point b) 25 { 26 return point(a.x-b.x,a.y-b.y); 27 } 28 int dcmp(double x) 29 { 30 if(fabs(x)<eps) return 0; 31 return x<0?-1:1; 32 } 33 double dis(point a,point b) 34 { 35 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y)); 36 } 37 double cross(point a,point b) 38 { 39 return a.x*b.y-a.y*b.x; 40 } 41 double mul(point p0,point p1,point p2) 42 { 43 return cross(p1-p0,p2-p0); 44 } 45 double dis(point a) 46 { 47 return sqrt(a.x*a.x+a.y*a.y); 48 } 49 bool cmp(point a,point b) 50 { 51 if(dcmp(mul(p[0],a,b))==0) 52 return dis(a-p[0])<dis(b-p[0]); 53 else 54 return dcmp(mul(p[0],a,b))>0; 55 } 56 int Graham(int n) 57 { 58 int i,k = 0,top; 59 point tmp; 60 for(i = 0 ; i < n; i++) 61 { 62 if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x)) 63 k = i; 64 } 65 if(k!=0) 66 { 67 tmp = p[0]; 68 p[0] = p[k]; 69 p[k] = tmp; 70 } 71 sort(p+1,p+n,cmp); 72 ch[0] = p[0]; 73 ch[1] = p[1]; 74 top = 1; 75 for(i = 2; i < n ; i++) 76 { 77 while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<0) 78 top--; 79 top++; 80 ch[top] = p[i]; 81 } 82 return top; 83 } 84 double distancetoline(point p,point a,point b) 85 { 86 point v1 = b-a,v2 = p-a; 87 return fabs(cross(v1,v2))/dis(v1); 88 } 89 int main() 90 { 91 int i,j; 92 int n,kk=0; 93 while(scanf("%d",&n)&&n) 94 { 95 for(i = 0 ; i < n; i++) 96 scanf("%lf%lf",&p[i].x,&p[i].y); 97 int m = Graham(n); 98 ch[m+1] = ch[0]; 99 double ans = INF; 100 for(i = 0 ; i <= m; i++) 101 { 102 double ts = 0; 103 for(j = 0 ; j <= m ; j++) 104 ts = max(ts,distancetoline(ch[j],ch[i],ch[i+1])); 105 ans = min(ans,ts); 106 } 107 ans = ceil(ans*100)/100; 108 printf("Case %d: %.2lf ",++kk,ans); 109 } 110 return 0; 111 }