UVA 270 Lining Up 共线点 暴力

题意：给出几个点的位置，问一条直线最多能连过几个点。

只要枚举每两个点组成的直线，然后找直线上的点数，更新最大值即可。

我这样做过于暴力，2.7s让人心惊肉跳。。。应该还能继续剪枝的，同一直线找过之后就可以剪掉了。

代码：

 

 /*

 *   Author:        illuz <[email protected]>

 *   Blog:          http://blog.csdn.net/hcbbt

 *   File:          uva270.cpp

 *   Lauguage:      C/C++

 *   Create Date:   2013-08-25 10:54:55

 *   Descripton:    UVA 270	 Lining Up, greed, enumeration, brute force

 */

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cmath>

#include <iostream>

#include <list>

#include <vector>

#include <map>

#include <set>

#include <deque>

#include <queue>

#include <stack>

#include <utility>

#include <algorithm>

using namespace std;

#define rep(i, n) for (int i = 0; i < (n); i++)

#define repu(i, a, b) for (int i = (a); i < (b); i++)

#define repf(i, a, b) for (int i = (a); i <= (b); i++)

#define repd(i, a, b) for (int i = (a); i >= (b); i--)

#define swap(a, b) {int t = a; a = b; b = t;}

#define mc(a) memset(a, 0, sizeof(a))

#define ms(a, i) memset(a, i, sizeof(a))

#define sqr(x) ((x) * (x))

#define FI(i, x) for (typeof((x).begin()) i = (x).begin(); i != (x).end(); i++)

typedef long long LL;

typedef unsigned long long ULL;



/****** TEMPLATE ENDS ******/



const int MAXN = 710;

int cas, n;

char str[100];

pair<int, int> p[MAXN];



int main() {

	scanf("%d\n", &cas);

	rep(c, cas) {

		n = 0;

		while (gets(str) && strlen(str)) {

//			puts(str);

			sscanf(str, "%d%d", &p[n].first, &p[n].second);

			n++;

		}

		int Max = 1, sum;

		rep(i, n) repu(j, i + 1, n) {

			sum = 2;

			int tx = p[i].first - p[j].first, ty = p[i].second - p[j].second;

			rep(k, n) {

				if (k == i || k == j) continue;

				if ((p[k].first - p[i].first) * ty == (p[k].second - p[i].second) * tx)

					sum++;

			}

			if (sum > Max)

				Max = sum;

		}

		if (c) printf("\n");

		printf("%d\n", Max);

	}

	return 0;

}