poj1118(Lining Up)

 题目大意是这样:给出n个点的坐标，判断在这些点中最多有多少个点共线。。。

我开始的思路是用暴力，找出前两个点的连线然后再从生下的点中找出斜率相同的点记录并且统计个数，最后找出统计的个数中最大的，时间复杂度是O(n^3)数据最大是700，所以我开始希望可以险过，可是结果没过，所以我就又换了一种思路重新思考，对于从每个起点开始，记录每种斜率的个数，最后返回最大值，这种算法只用了O(n^2)

Lining Up

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 15964		Accepted: 5003

Description

"How am I ever going to solve this problem?" said the pilot.

Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number?


Your program has to be efficient!

Input

Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.

Output

output one integer for each input case ,representing the largest number of points that all lie on one line.

Sample Input

5
1 1
2 2
3 3
9 10
10 11
0

Sample Output

3

 

***********************、、、、、、超时代码、、、、、、、、、、、、、***********

#include
#include

using namespace std;

struct point{
    int x,y;
}dian[750];
int n;

double xielv(point p1, point p2){
    return (double)(p1.y - p2.y)/(double)(p1.x - p2.x);
}
int cal_xielv(){
    int max = 0;
    int geshu;
    for(int i = 0; i < n - 1;i++){
        if(n - i < max){break;}
        for(int j =  i + 1; j < n; j++){
            geshu = 2;
            if(n - j < max){break;}
            for(int k = j + 1; k < n; k++){
                if(dian[i].x == dian[j].x){
                    if(dian[j].x == dian[k].x){
                        geshu++;
                    }
                }
                else if(xielv(dian[i],dian[j]) == xielv(dian[j], dian[k])){
                    geshu++;
                }
            }
            if(geshu > max)  max = geshu;
        }
    }
    return max;
}
int main(){
    while(scanf("%d", &n)&&n!=0){
        for(int i = 0; i < n; i++){
            scanf("%d%d", &dian[i].x, &dian[i].y);
        }
        printf("%d\n", cal_xielv());
    }
    return 0;
}

以下为AC代码：

708K

313MS

开始的时候我把max初始化为0，造成了一个很难查找的错误，因此接着就很长一串wa，真是悲剧啊，下次注意了、、、、、、、

#include
#include
#include
using namespace std;

struct point{
    int x,y;
}dian[750];
int n;
const int INF = 0x7fffffff;
double lie_xie[750];
double xielv(point p1, point p2){
    return (double)(p1.y - p2.y)/(double)(p1.x - p2.x);
}

int cmp(const void *a,const void *b){
    return *(double *)a > *(double *)b ? 1 : -1;
}
int cal_xielv(){
    int max = 2;
    int geshu;
    for(int i = 0; i < n;i++){
        geshu = 2;
        int xienum = 0;//斜率个数
        for(int j = i + 1; j < n; j++){
            if(dian[i].x == dian[j].x){
                lie_xie[xienum++] = INF;
            }
            else {
                lie_xie[xienum++] = xielv(dian[i], dian[j]);
            }
        }
        qsort(lie_xie,xienum,sizeof(lie_xie[0]),cmp);
        for(int k = 0; k < xienum - 1;k++){
            if(lie_xie[k] == lie_xie[k + 1]){
                ++geshu;
                if(geshu > max)  max  = geshu;
            }
            else geshu = 2;
        }
    }
    return max;
}
int main(){
    while(scanf("%d", &n)&&n!=0){
        for(int i = 0; i < n; i++){
            scanf("%d%d", &dian[i].x, &dian[i].y);
        }
        printf("%d\n", cal_xielv());
    }
    return 0;
}