poj2002Squares(hash)
http://poj.org/problem?id=2002
刚开始hash的两点 TLE了 又改为一点 已知对角两点 求令两点
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1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 #define MOD 977777 6 int head[1000010],v[1000010],next[1000010],num[1000010][2],t,f[1010][1010]; 7 struct node 8 { 9 int x,y; 10 }co[1010]; 11 int judge(int s,struct node x) 12 { 13 int i; 14 for(i = head[s] ; i!= -1 ; i = next[i]) 15 if(co[v[i]].x==x.x&&co[v[i]].y==x.y) 16 return v[i]; 17 return 0; 18 } 19 int DoneSq(struct node a, struct node c)//求令两点 网上搜的公式 20 { 21 struct node b, d; 22 double x,y,mx, my,tx,ty; 23 int s1,s2,i; 24 mx = (a.x+c.x)/2.0, my = (a.y+c.y)/2.0; 25 x = a.x - mx; y = a.y - my; 26 tx = -y + mx; ty = x + my; 27 if((tx-(int)tx)!=0||(ty-(int)ty)!=0) 28 return 0; 29 b.x = tx;b.y = ty; 30 x = c.x - mx; y = c.y - my; 31 tx = - y + mx; ty = x + my; 32 if((tx-(int)tx)!=0||(ty-(int)ty)!=0) 33 return 0; 34 d.x = tx;d.y=ty; 35 s1 = b.x*977+b.y; s1 = s1%MOD; 36 if(s1<0) s1+=MOD; 37 s2 = d.x*977+d.y; s2 = s2%MOD; 38 if(s2<0) s2+=MOD; 39 if(judge(s1,b)&&judge(s2,d)) 40 { 41 f[judge(s1,b)][judge(s2,d)] = 1; 42 f[judge(s2,d)][judge(s1,b)] = 1; 43 return 1; 44 } 45 return 0; 46 47 } 48 void init() 49 { 50 t = 0; 51 memset(head,-1,sizeof(head)); 52 } 53 void add(int a,int b) 54 { 55 next[t] = head[a]; 56 head[a] = t; 57 v[t] = b; 58 t++; 59 } 60 int main() 61 { 62 int i,j,k,n,g,s,numm; 63 while(cin>>n) 64 { 65 if(n==0) 66 break; 67 init(); 68 memset(f,0,sizeof(f)); 69 g = 0;numm=0; 70 for(i = 1; i <= n ; i++) 71 { 72 cin>>co[i].x>>co[i].y; 73 s = co[i].x*977+co[i].y; 74 s = s%MOD; 75 if(s<0) 76 s+=MOD; 77 add(s,i); 78 } 79 for(i = 1; i <= n ; i++) 80 for(j = i+1 ; j <= n ; j++) 81 { 82 if(!f[i][j]) 83 numm+=DoneSq(co[i],co[j]); 84 } 85 cout<<numm<<endl; 86 } 87 return 0; 88 }
