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  • POJ1548 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 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/2868251.html
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