题目链接：http://poj.org/problem?id=2002
测试数据：
4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0
有两种解法，第一种用二分查找，第二种用到hash存储：
解法一：
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 int T; 8 struct TT 9 { 10 int x,y; 11 }node[1010]; 12 bool cmp(TT a,TT b) 13 { 14 if(a.x<b.x ||(a.x==b.x && a.y<b.y)) return true; 15 return false; 16 } 17 bool judge(int x,int y) 18 { 19 int L = 1,R =T; 20 while(L<=R) 21 { 22 int mid = (L+R)/2; 23 if(node[mid].x == x && node[mid].y == y) 24 { 25 return true; 26 } 27 if(x == node[mid].x ) //这个地方老是出错，就是当x值相同的时候，会有两种情况的，我只是考虑了一种 28 { 29 if(y>node[mid].y) L = mid +1; 30 else R = mid-1; 31 } 32 else if(x>node[mid].x) L = mid+1; 33 else R = mid-1; 34 } 35 return false; 36 } 37 int main() 38 { 39 int ans; 40 int x1,y1,x2,y2,mx,my; 41 while(scanf("%d",&T) && T) 42 { ans = 0; 43 for(int i =1;i<=T;i++) 44 scanf("%d %d",&node[i].x,&node[i].y); 45 sort(node+1,node+T+1,cmp);//乱了一下，开始从0，开始排序了 46 for(int i = 1; i<T; i++) 47 for(int j = i+1; j<=T; j++) 48 { 49 mx = node[j].x - node[i].x; 50 my = node[j].y - node[i].y; 51 x1 = node[i].x+my; 52 y1 = node[i].y-mx; 53 x2 = node[j].x+my; 54 y2 = node[j].y-mx; 55 if( judge(x1,y1) && judge(x2,y2)) 56 { 57 ans++; 58 } 59 } 60 printf("%d ",ans/2);//因为每次都有两个重复； 61 } 62 return 0; 63 }
解法二：本来想着哈希简简单单就过了呢，咋说也比二分省时间吧，可是呢，整了两个小时才搞定，还发现了一个问题；有待于求证：结构体数组赋值时间小于数组赋值时间，
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 struct TT 8 { 9 int x,y,next; 10 }node[1010],hash[1010]; 11 const int N = 1100; 12 int vis[N][N]; 13 int T; 14 const int MOD = 10007; 15 int next[N],head[MOD],top = 1; 16 void creath(int x,int y) 17 { 18 int key = abs(x)%MOD; 19 hash[top].next = head[key];//这样写超时： next[top] = head[key]; 20 hash[top].x = x; 21 hash[top].y = y; 22 head[key] = top; 23 top++; 24 } 25 bool cmp(TT a,TT b) 26 { 27 if(a.x<b.x || (a.x == b.x && a.y<b.y)) return true; 28 return false; 29 } 30 bool judge(int x,int y) 31 { 32 int key = abs(x)%MOD; 33 int ans = 0; 34 for(int i = head[key];i>=0;i = hash[i].next)//换成 i = next[i] 超时 35 { 36 if(hash[i].x==x && hash[i].y == y) 37 return true; 38 } 39 return false; 40 } 41 int main() 42 { 43 int ans; 44 int x1,y1,x2,y2,mx,my; 45 while(scanf("%d",&T) && T) 46 { memset(head,-1,sizeof(head)); 47 ans = 0; 48 for(int i =1;i<=T;i++) 49 { 50 scanf("%d %d",&node[i].x,&node[i].y); 51 creath(node[i].x,node[i].y); 52 } 53 sort(node+1,node+1+T,cmp); 54 for(int i = 1; i<T; i++) 55 for(int j = i+1; j<=T; j++) 56 { 57 mx = node[j].x - node[i].x; 58 my = node[j].y - node[i].y; 59 x1 = node[i].x+my; 60 y1 = node[i].y-mx; 61 x2 = node[j].x+my; 62 y2 = node[j].y-mx; 63 if( judge(x1,y1) && judge(x2,y2)) 64 { 65 ans++; 66 } 67 } 68 printf("%d ",ans/2); 69 } 70 return 0; 71 }