二分+计算几何/半平面交
半平面交的学习戳这里：http://blog.csdn.net/accry/article/details/6070621
然而这题是要二分长度r……用每条直线的距离为r的平行线来截“凸包”
做平行线的方法是：对于向量(x,y)，与它垂直的向量有：(y,-x)和(-y,x)，然后再转化成单位向量，再乘以 r ，就得到了转移向量tmp，设直线上两点为A(x1,y1),B(x2,y2)，用这个方法就找到了直线AB的平行线CD，其中C=A+tmp,D=b+tmp;
然而我傻逼的搞错了方向……结果平行线做成了远离凸包的那一条……导致答案一直增大QAQ
剩下的就没啥了，看是否存在这么一块半平面的交即可。
1 Source Code 2 Problem: 3525 User: sdfzyhy 3 Memory: 672K Time: 16MS 4 Language: G++ Result: Accepted 5 6 Source Code 7 8 //POJ 3525 9 #include<cmath> 10 #include<cstdio> 11 #include<cstring> 12 #include<cstdlib> 13 #include<iostream> 14 #include<algorithm> 15 #define rep(i,n) for(int i=0;i<n;++i) 16 #define F(i,j,n) for(int i=j;i<=n;++i) 17 #define D(i,j,n) for(int i=j;i>=n;--i) 18 using namespace std; 19 typedef long long LL; 20 inline int getint(){ 21 int r=1,v=0; char ch=getchar(); 22 for(;!isdigit(ch);ch=getchar()) if(ch=='-')r=-1; 23 for(; isdigit(ch);ch=getchar()) v=v*10+ch-'0'; 24 return r*v; 25 } 26 const int N=110; 27 const double eps=1e-6,INF=1e8; 28 /*******************template********************/ 29 30 31 int n; 32 struct Poi{ 33 double x,y; 34 Poi(){} 35 Poi(double x,double y):x(x),y(y){} 36 void read(){scanf("%lf%lf",&x,&y);} 37 void out(){printf("%f %f ",x,y);} 38 }p[N],tp[N],s[N],o; 39 typedef Poi Vec; 40 Vec operator - (const Poi &a,const Poi &b){return Vec(a.x-b.x,a.y-b.y);} 41 Vec operator + (const Poi &a,const Vec &b){return Poi(a.x+b.x,a.y+b.y);} 42 Vec operator * (const double &k,const Vec a){return Vec(a.x*k,a.y*k);} 43 inline double Cross(const Vec &a,const Vec &b){return a.x*b.y-a.y*b.x;} 44 inline double Dot(const Vec &a,const Vec &b){return a.x*b.x+a.y*b.y;} 45 inline double Len(const Vec &a){return sqrt(Dot(a,a));} 46 inline int dcmp(double x){return x > eps ? 1 : x < eps ? -1 : 0;} 47 double getarea(Poi *a,int n){ 48 double ans=0.0; 49 F(i,1,n) ans+=Cross(a[i]-o,a[i+1]-o); 50 return ans*0.5; 51 } 52 Poi getpoint (const Poi &a,const Poi &b,const Poi &c,const Poi &d){ 53 Poi ans,tmp=b-a; 54 double k1=Cross(d-a,c-a),k2=Cross(c-b,d-b); 55 ans=a+(k1/(k1+k2))*tmp; 56 return ans; 57 } 58 inline void get_parallel(Poi &a,Poi &b,double r,Poi &c,Poi &d){ 59 // Poi tmp(b.y-a.y,a.x-b.x); 60 Poi tmp(a.y-b.y,b.x-a.x); 61 tmp=(r/Len(tmp))*tmp; 62 c=a+tmp; d=b+tmp; 63 // a.out(); b.out(); c.out(); d.out(); 64 // puts(""); 65 } 66 void init(){ 67 F(i,1,n) p[i].read(); 68 p[n+1]=p[1]; 69 } 70 bool check(double r){ 71 tp[1]=tp[5]=Poi(-INF,-INF); 72 tp[2]=Poi(INF,-INF); 73 tp[3]=Poi(INF,INF); 74 tp[4]=Poi(-INF,INF); 75 int num=4,size=0; 76 F(i,1,n){ 77 size=0; 78 F(j,1,num){ 79 Poi t1(0,0),t2(0,0); 80 get_parallel(p[i],p[i+1],r,t1,t2); 81 if (dcmp(Cross(t2-t1,tp[j]-t1))>=0) 82 s[++size]=tp[j]; 83 if (dcmp(Cross(t2-t1,tp[j]-t1)*Cross(t2-t1,tp[j+1]-t1))<0) 84 s[++size]=getpoint(t1,t2,tp[j],tp[j+1]); 85 } 86 s[size+1]=s[1]; 87 F(j,1,size+1) tp[j]=s[j]; 88 num=size; 89 // printf("num=%d ",num); 90 // F(j,1,num) tp[j].out(); 91 } 92 // printf("n=%d num=%d size=%d ",n,num,size); 93 if (num>0) return 1; 94 else return 0; 95 } 96 int main(){ 97 #ifndef ONLINE_JUDGE 98 freopen("3525.in","r",stdin); 99 freopen("3525.out","w",stdout); 100 #endif 101 while(scanf("%d",&n)!=EOF && n){ 102 init(); 103 double l=0.0,r=1e7,mid; 104 while(r-l>eps){ 105 mid=(l+r)/2; 106 // printf("mid=%f ",mid); 107 if (check(mid)) l=mid; 108 else r=mid; 109 } 110 printf("%f ",l); 111 } 112 return 0; 113 }
| Time Limit: 5000MS | Memory Limit: 65536K | |||
| Total Submissions: 4183 | Accepted: 1956 | Special Judge | ||
Description
The main land of Japan called Honshu is an island surrounded by the sea. In such an island, it is natural to ask a question: “Where is the most distant point from the sea?” The answer to this question for Honshu was found in 1996. The most distant point is located in former Usuda Town, Nagano Prefecture, whose distance from the sea is 114.86 km.
In this problem, you are asked to write a program which, given a map of an island, finds the most distant point from the sea in the island, and reports its distance from the sea. In order to simplify the problem, we only consider maps representable by convex polygons.
Input
The input consists of multiple datasets. Each dataset represents a map of an island, which is a convex polygon. The format of a dataset is as follows.
| n | ||
| x1 | y1 | |
| ⋮ | ||
| xn | yn |
Every input item in a dataset is a non-negative integer. Two input items in a line are separated by a space.
n in the first line is the number of vertices of the polygon, satisfying 3 ≤ n ≤ 100. Subsequent n lines are the x- and y-coordinates of the n vertices. Line segments (xi, yi)–(xi+1, yi+1) (1 ≤ i ≤ n − 1) and the line segment (xn, yn)–(x1, y1) form the border of the polygon in counterclockwise order. That is, these line segments see the inside of the polygon in the left of their directions. All coordinate values are between 0 and 10000, inclusive.
You can assume that the polygon is simple, that is, its border never crosses or touches itself. As stated above, the given polygon is always a convex one.
The last dataset is followed by a line containing a single zero.
Output
For each dataset in the input, one line containing the distance of the most distant point from the sea should be output. An output line should not contain extra characters such as spaces. The answer should not have an error greater than 0.00001 (10−5). You may output any number of digits after the decimal point, provided that the above accuracy condition is satisfied.
Sample Input
4 0 0 10000 0 10000 10000 0 10000 3 0 0 10000 0 7000 1000 6 0 40 100 20 250 40 250 70 100 90 0 70 3 0 0 10000 10000 5000 5001 0
Sample Output
5000.000000 494.233641 34.542948 0.353553
Source