HDU 6055 Regular polygon
2017-08-07 08:55
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【题目链接】
http://acm.hdu.edu.cn/showproblem.php?pid=6055
http://acm.hdu.edu.cn/showproblem.php?pid=6055
题目意思
给定n个点问你可以形成多少个正多边形。解题思路
因为给定点为整数,所以只能形成正4边形,因此从上往下,从左往右。任意两个点两两判断剩余其余两点是否在给定点中,最后答案除2,因为ab和ba重复计算代码部分
#include <bits/stdc++.h> using namespace std; const int maxn=510; struct Node ///存储点 { int x,y; bool operator<(const Node &rhs)const ///排序从上往下从左往右 { if(x == rhs.x) return y < rhs.y; else return x < rhs.x; } } nodes[maxn]; int main() { int n; while(~scanf("%d",&n)) { int ans=0;///正方形个数 for(int i=0; i<n; ++i) scanf("%d%d",&nodes[i].x,&nodes[i].y); sort(nodes,nodes+n); for(int i=0; i<n; ++i) for(int j=i+1; j<n; ++j) { Node tmp; ///tmp作为正方形的第3或4个点 ///特殊方法来判断第三个和第四个点的位置(自己画画图就懂了) tmp.x=nodes[i].x+nodes[i].y-nodes[j].y; tmp.y=nodes[i].y+nodes[j].x-nodes[i].x; if(!binary_search(nodes,nodes+n,tmp)) continue; ///二分查找是否存在这个点 tmp.x=nodes[j].x+nodes[i].y-nodes[j].y; tmp.y=nodes[j].y+nodes[j].x-nodes[i].x; if(!binary_search(nodes,nodes+n,tmp)) continue; ++ans; } ///在两个点为一个边的时候判断了两次,所以结果应该除二 printf("%d\n",ans/2); } return 0; }
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