HDU6055 Regular polygon 2017 Multi-University Training Contest
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Regular polygon
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2924 Accepted Submission(s): 685
Problem Description
On a two-dimensional plane, give you n integer points. Your task is to figure out how many different regular polygon these points can make.
Input
The input file consists of several test cases. Each case the first line is a numbers N (N <= 500). The next N lines ,each line contain two number Xi and Yi(-100 <= xi,yi <= 100), means the points’ position.(the data assures no two points share the same position.)
Output
For each case, output a number means how many different regular polygon these points can make.
Sample Input
40 00 11 01 160 00 11 01 12 02 1
Sample Output
12
题目意思:给你n个整数点,问这n个整数点内能构成多少个正多边形。
官方解释:题意,二维平面上给N个整数点,问能构成多少个不同的正多边形。 题解:容易得知只有正四边形可以使得所有的顶点为整数点。(具体证明可参考杨景钦在2017的国家队论文) 所以正解即求出所有的正四边形个数。 枚举2个点,然后暴力判断另外2个点的位置是否存在。 复杂度 N*N*logN。
因为点的数量最多为500 所以可以考虑暴力求解。
涉及到的知识点有,哈希表,链式前向星存储,暴力求解。
链式前向星不懂的可以看这个博客,讲的很详细:http://blog.csdn.net/acdreamers/article/details/16902023
哈希表:http://www.cnblogs.com/dolphin0520/archive/2012/09/28/2700000.html
代码如下:
/*2017年7月28日16:34:20AC代码 */ #include<stdio.h>#include<string.h>#include<math.h>using namespace std; typedef long long ll;const ll mod=1010;const ll maxn=550;struct node{ll x;ll y;ll next;}A[maxn];ll X[maxn],Y[maxn];ll H[mod];ll node,n,ans;/*对每个坐标进行存储利用哈希表 ,将二维坐标通过 key=(x*x+y*y)%mod这种计算方式映射到一维这里也利用了链式前向星的存储方式,*/ void insert(int x,int y){int key=(x*x+y*y)%mod;A[node].x=x;A[node].y=y;A[node].next=H[key];H[key]=node++;//printf("H[%d]=%d\n",key,node-1);//大家要是不懂这种存储方式,可以把这两句printf()加上,看看这几个点在H[] 和 A[] 中到底是怎么存储的//printf("A[%d].x=%d A[%d].y=%d A[%d].next=%d\n",node-1,A[node-1].x,node-1,A[node-1].y,node-1,A[node-1].next);}bool search(int x,int y){int key=(x*x+y*y)%mod;/*链式前向星的遍历方式 */ for(int k=H[key];k!=-1;k=A[k].next){if(A[k].x==x&&A[k].y==y) return true;}return false;}int main(){while(scanf("%d",&n)==1){node=0;ans=0;memset(H,-1,sizeof(H));for(int i=1;i<=n;i++){scanf("%d%d",&X[i],&Y[i]);insert(X[i],Y[i]);} /*for(int i=0;i<node;i++){printf("A[%d].x=%d A[%d].y=%d A[%d].next=%d\n",i,A[i].x,i,A[i].y,i,A[i].next);}*/ for(int i=1;i<=n;i++){for(int j=i+1;j<=n;j++){int x1,x2,y1,y2;x1=X[i]-(Y[i]-Y[j]);y1=Y[i]+(X[i]-X[j]);x2=X[j]-(Y[i]-Y[j]);y2=Y[j]+(X[i]-X[j]);if(search(x1,y1)&&search(x2,y2))ans++;x1=X[i]+(Y[i]-Y[j]);y1=Y[i]-(X[i]-X[j]);x2=X[j]+(Y[i]-Y[j]);y2=Y[j]-(X[i]-X[j]);if(search(x1,y1)&&search(x2,y2))ans++;}}/*考虑到每一条边都可以作为一个正四边形的四条边来算,所以最后的结果要除以4 */ printf("%d\n",ans/4);}return 0;}
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