hdu 5020(斜率的表示+STL)

Revenge of Collinearity

Time Limit: 8000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 612    Accepted Submission(s): 207

Problem Description

In geometry, collinearity is a property of a set of points, specifically, the property of lying on a single line. A set of points with this property is said to be collinear (often misspelled as colinear).
---Wikipedia

Today, Collinearity takes revenge on you. Given a set of N points in two-dimensional coordinate system, you have to find how many set of <P_i, P_j, P_k> from these N points are collinear. Note that <P_i, P_j, P_k> cannot contains same point, and <P_i, P_j, P_k> and <P_i, P_k, P_j> are considered as the same set, i.e. the order in the set doesn’t matter.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with an integer N, following N lines, each line contains two integers Xi and Yi, describing a point.

[Technical Specification]
1. 1 <= T <= 33
2. 3 <= N <= 1 000
3. -1 000 000 000 <= Xi, Yi <= 1 000 000 000, and no two points are identical.
4. The ratio of test cases with N > 100 is less than 25%.

 

Output

For each query, output the number of three points set which are collinear.

 

Sample Input

2 3 1 1 2 2 3 3 4 0 0 1 0 0 1 1 1

 

Sample Output

1 0

 

Source

BestCoder Round #10

 

题意:平面上有n个点,统计有多少个三个点在一条直线上（注意:(1,2,3)和(1,3,2)相同）。

题解:这题O（n^3）会炸。。所以想办法优化，然后就想到斜率。但是斜率不存在怎么办？？

1.然后看了别人的代码，求出 y1 - y2 与 x1 - x2 的最大公约数d，斜率可以直接用(（y1-y2)/d,(x1-x2)/d)进行唯一的表示。

2。对所有点进行排序，先按x坐标,然后再按y坐标排序。接着对于第 i 个点,我们找到 p (p>=2)个点与其共线,那么总共有p*(p-2)种可能的组合方法。

3.用pair 和 map 将条件 1，2联系起来。map进行计数,pair用来存斜率。

4.好巧妙的一个题。

pair是可以直接根据里面的元素进行判断是否相等的.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<int,pair<int,int> > piii;
int main()
{
    pii a(1,2);
    pii b(1,2);
    printf("%d
",a==b);
    piii c(1,make_pair(1,2));
    piii d(1,make_pair(1,2));
    printf("%d
",c==d);
    return 0;
}
/*1 1*/

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
const int N = 1005;
struct Point{
    int x,y;
}p[N];
int cmp(Point a,Point b){
    if(a.x==b.x) return a.y<b.y;
    return a.x<b.y;
}
int gcd(int a,int b){
    return b==0?a:gcd(b,a%b);
}
map<pair<int,int>,int> mp;
map<pair<int,int>,int>::iterator it;
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--){
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&p[i].x,&p[i].y);
        }
        int ans = 0;
        for(int i=1;i<=n;i++){
            mp.clear();
            for(int j=i+1;j<=n;j++){
                int x = p[j].x - p[i].x;
                int y = p[j].y - p[i].y;
                int d = gcd(x,y);
                x/=d,y/=d;
                mp[make_pair(x,y)]++;
            }
            for(it = mp.begin();it!=mp.end();it++){
                if(it->second>=2){
                    int t = it->second;
                    ans+=t*(t-1)/2;
                }
            }
        }
        printf("%d
",ans);
    }
    return 0;
}