http://codeforces.com/contest/1017/problem/E
凸包模板+kmp
1 #include <cstdio> 2 #include <cstdlib> 3 #include <cmath> 4 #include <cstring> 5 #include <time.h> 6 #include <string> 7 #include <set> 8 #include <map> 9 #include <list> 10 #include <stack> 11 #include <queue> 12 #include <vector> 13 #include <bitset> 14 #include <ext/rope> 15 #include <algorithm> 16 #include <iostream> 17 using namespace std; 18 #define ll long long 19 #define minv 1e-10 20 #define inf 1e9 21 #define pi 3.1415926536 22 #define E 2.7182818284 23 const ll mod=1e9+7;//998244353 24 const int maxn=1e5+10; 25 26 struct node 27 { 28 ll x,y; 29 }a[maxn],b[maxn],e[maxn],f[maxn],point; 30 31 int p[maxn*6]; 32 double c[maxn*6],d[maxn*6]; 33 34 ///result only 1/0 35 int cmp(node a,node b) 36 { 37 ll s=(a.y-point.y)*(b.x-point.x) - (b.y-point.y)*(a.x-point.x); 38 if (s<0) 39 return 1; 40 else if (s>0) 41 return 0; 42 else 43 return pow(a.y-point.y,2)+pow(a.x-point.x,2) < pow(b.y-point.y,2)+pow(b.x-point.x,2); 44 } 45 46 void work(node a[],double c[],node e[],int n,int* z) 47 { 48 int i,g; 49 for (i=1;i<=n;i++) 50 { 51 scanf("%lld%lld",&a[i].x,&a[i].y); 52 if (a[1].x>a[i].x || (a[1].x==a[i].x && a[1].y>a[i].y)) 53 swap(a[i],a[1]); 54 } 55 point=a[1]; 56 sort(a+2,a+n+1,cmp); 57 58 e[1].x=a[1].x,e[1].y=a[1].y; 59 e[2].x=a[2].x,e[2].y=a[2].y; 60 g=2; 61 for (i=3;i<=n;i++) 62 { 63 while (g>1 && (a[i].x-e[g-1].x)*(e[g].y-e[g-1].y)-(e[g].x-e[g-1].x)*(a[i].y-e[g-1].y)>=0) 64 g--; 65 g++; 66 e[g].x=a[i].x,e[g].y=a[i].y; 67 } 68 e[g+1]=e[1]; 69 e[0]=e[g]; 70 71 for (i=1;i<=g;i++) 72 { 73 //len^2 cos(angle) 74 c[i*2-1]=(double)(pow(e[i+1].x-e[i].x,2) + pow(e[i+1].y-e[i].y,2)); 75 76 ll b1=pow(e[i+1].y-e[i].y,2)+pow(e[i+1].x-e[i].x,2), 77 b2=pow(e[i-1].y-e[i].y,2)+pow(e[i-1].x-e[i].x,2), 78 b3=pow(e[i+1].y-e[i-1].y,2)+pow(e[i+1].x-e[i-1].x,2); 79 80 c[i*2]=1.0*(b1+b2-b3)/2/sqrt(b1)/sqrt(b2); 81 } 82 (*z)=g*2; 83 } 84 85 int main() 86 { 87 int n,m,g1,g2,i,j; 88 scanf("%d%d",&n,&m); 89 work(a,c,e,n,&g1); 90 work(b,d,f,m,&g2); 91 if (g1!=g2) 92 { 93 printf("NO"); 94 return 0; 95 } 96 for (i=1;i<=g2;i++) 97 d[i+g2]=d[i]; 98 g2<<=1; 99 100 p[1]=0; 101 j=0; 102 for (i=2;i<=g1;i++) 103 { 104 while (j>0 && fabs(c[j+1]-c[i])>minv) 105 j=p[j]; 106 if (fabs(c[j+1]-c[i])<minv) 107 j++; 108 p[i]=j; 109 } 110 111 j=0; 112 for (i=1;i<=g2;i++) 113 { 114 while (j>0 && fabs(c[j+1]-d[i])>minv) 115 j=p[j]; 116 if (fabs(c[j+1]-d[i])<minv) 117 j++; 118 if (j==g1) 119 { 120 printf("YES"); 121 return 0; 122 } 123 } 124 printf("NO"); 125 return 0; 126 } 127 /* 128 6 5 129 1 100000000 130 2 100000000 131 3 100000000 132 4 100000000 133 5 100000000 134 6 1 135 136 2 100000000 137 6 1 138 3 100000000 139 1 100000000 140 5 100000000 141 */