poj_2002
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源链接:http://poj.org/problem?id=2002
题目大意就是给出n个二维平面上的坐标,求出这n个坐标最多能够构成多少个正方形。
通过n最大为1000可以推测,应该是n^2左右的效率,所以我们先枚举两个点,然后通过这两个点再求出另外两个点的坐标,在输入的集合中查找是否存在这两个点,如果存在的话就可以构成一个正方形。
但是枚举也有一定的方法,我们不能随便枚举,因为有很多种形状的正方形,我们也不知道这两个点是正方形的哪两个点。首先,我们要先将所有的点按照x,y的大小排好序,然后遍历的时候取a[i]和a[j]两个点,且i<j,即a[i].x<a[j].x || a[i].x==a[j].x&&a[i].y<a[j].y。这样子我们就可以自定义是哪两个点:
比如这么一个正方形,我们已知1,2两个点的坐标为(a1,a2),(b1,b2),那么我们怎么求得3,4两个点的坐标那?其实正方形有它的特殊性。
我们可以看出,上下做的两个三角形是全等的,所以说对于点3,点3的坐标为(a1+(b2-a2),a2-(b1-a1)),同理,点4的坐标为(b1+(b2-a2),b2-(b1-a1))。知道了这两个点,就可以去找了。
但是如果会问,那么如果遍历到的是1,3怎么办?其实不用管,因为这个是按从小到大来遍历的,那么1,2两个点肯定会被遍历到,就不用管1,3了。还有一点,如果我们遍历到的是2,4两个顶点,那么计算1,3两个定点的时候,计算公式是和知道1,3一样的。所以,对于每个四边形,我们都遍历了两遍,最后除以2即可。
对于查找的算法,我们可以用哈希或者二分即可。
先是哈希:
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,t;#define Mod 1000000007#define N 110#define M 1000100struct Squares{int x,y;bool operator <(const Squares &s) const{return (x<s.x || (x==s.x && y<s.y));}}a[1010];int hash[M];int next[1010];bool check(int x,int y){int key = abs(x+y);int id = hash[key];while(id != -1){if(a[id].x == x && a[id].y == y)return true;id = next[id];}return false;}int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif while(sf(n)!=EOF && n){ for(int i=0;i<n;i++) sfd(a[i].x,a[i].y); memset(hash,-1,sizeof hash); memset(next,-1,sizeof next); sort(a,a+n); for(int i=0;i<n;i++){ int tmp = abs(a[i].x+a[i].y); //用两个坐标的绝对值和作为哈希值 next[i] = hash[tmp];<span style="white-space:pre"></span>//这个是模仿链式向前星的 hash[tmp] = i; } int ans = 0; for(int i=0;i<n;i++){ for(int j=i+1;j<n;j++){ int py = a[j].y-a[i].y; int px = a[j].x-a[i].x; int xx = a[i].x+py; int yy = a[i].y-px; if(!check(xx,yy))<span style="white-space:pre"></span>// continue; xx = a[j].x+py; yy = a[j].y-px; if(!check(xx,yy))<span style="white-space:pre"></span>// continue; ans++; } } printf("%d\n",ans/2); }return 0;}
然后是二分法:
#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m,t;#define Mod 1000000007#define N 110#define M 1000100struct Squares{int x,y;bool operator <(const Squares &s) const{return (x<s.x || (x==s.x && y<s.y));}}a[1010];bool cmp(int x,int y,Squares node){if(x<node.x)return true;if(x==node.x && y<node.y)return true;return false;}bool check(int x,int y){int l=0,r = n;int mid;while(l<=r){mid = (l+r)>>1;if(a[mid].x == x &&a[mid].y == y)return true;if(cmp(x,y,a[mid]))r = mid-1;elsel = mid+1;}return false;}int main(){#ifndef ONLINE_JUDGE freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);#endif while(sf(n)!=EOF && n){ for(int i=0;i<n;i++) sfd(a[i].x,a[i].y); sort(a,a+n); int ans = 0; for(int i=0;i<n;i++){ for(int j=i+1;j<n;j++){ int py = a[j].y-a[i].y; int px = a[j].x-a[i].x; int xx = a[i].x+py; int yy = a[i].y-px; if(!check(xx,yy)) continue; xx = a[j].x+py; yy = a[j].y-px; if(!check(xx,yy)) continue; ans++; } } printf("%d\n",ans/2); }return 0;}