POJ-1548 A Round Peg in a Ground Hole 凸多边形
题目链接：http://poj.org/problem?id=1584
首先判断是否为凸多边形，叉积判断即可，然后判断点是否在多边形内，先用叉积然后点到直线距离。
1 //STATUS:C++_AC_0MS_192KB 2 #include<stdio.h> 3 #include<stdlib.h> 4 #include<string.h> 5 #include<math.h> 6 #include<iostream> 7 #include<string> 8 #include<algorithm> 9 #include<vector> 10 #include<queue> 11 #include<stack> 12 using namespace std; 13 #define LL __int64 14 #define pii pair<int,int> 15 #define Max(a,b) ((a)>(b)?(a):(b)) 16 #define Min(a,b) ((a)<(b)?(a):(b)) 17 #define mem(a,b) memset(a,b,sizeof(a)) 18 #define lson l,mid,rt<<1 19 #define rson mid+1,r,rt<<1|1 20 const int N=210,M=1000000,INF=0x3f3f3f3f,MOD=1999997; 21 const LL LLNF=0x3f3f3f3f3f3f3f3fLL; 22 const double DNF=100000000; 23 24 struct Node{ 25 double x,y; 26 }nod[N],peg; 27 int n; 28 double pr; 29 30 double disln(Node &l1,Node &l2,Node &o){ 31 double A,B,C; 32 A=-(l1.y-l2.y); 33 B=l1.x-l2.x; 34 C=-A*l1.x-B*l1.y; 35 return fabs(A*o.x+B*o.y+C)/sqrt(A*A+B*B); 36 } 37 38 inline void getr(Node &r,Node *a) 39 { 40 r.x=a[1].x-a[0].x; 41 r.y=a[1].y-a[0].y; 42 } 43 44 int ispro(Node *a) 45 { 46 int i,ok; 47 double ini; 48 Node r1,r2; 49 getr(r1,a);getr(r2,a+1); 50 ini=r1.x*r2.y-r2.x*r1.y; 51 for(i=2;i<=n;i++){ 52 r1=r2; 53 getr(r2,a+i); 54 if((r1.x*r2.y-r2.x*r1.y)*ini<0)return 0; 55 } 56 57 return 1; 58 } 59 60 int isconcir(Node *a) 61 { 62 int i,j; 63 Node r1,r2; 64 double ini; 65 for(i=0;i<n;i++) 66 if(disln(a[i],a[i+1],peg)<pr)return 0; 67 getr(r1,a); 68 r2.x=peg.x-a[0].x;r2.y=peg.y-a[0].y; 69 ini=r1.x*r2.y-r2.x*r1.y; 70 for(i=1;i<n;i++){ 71 getr(r1,a+i); 72 r2.x=peg.x-a[i].x;r2.y=peg.y-a[i].y; 73 if((r1.x*r2.y-r2.x*r1.y)*ini<0)return 0; 74 } 75 return 1; 76 } 77 78 int main() 79 { 80 // freopen("in.txt","r",stdin); 81 int i,j; 82 while(~scanf("%d",&n) && n>2) 83 { 84 scanf("%lf%lf%lf",&pr,&peg.x,&peg.y); 85 for(i=0;i<n;i++){ 86 scanf("%lf%lf",&nod[i].x,&nod[i].y); 87 } 88 nod[n].x=nod[0].x;nod[n].y=nod[0].y; 89 nod[n+1].x=nod[1].x;nod[n+1].y=nod[1].y; 90 91 if(!ispro(nod))printf("HOLE IS ILL-FORMED\n"); 92 else if(isconcir(nod))printf("PEG WILL FIT\n"); 93 else printf("PEG WILL NOT FIT\n"); 94 } 95 return 0; 96 }
