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Description
| Lining Up |
``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!
The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank
line between two consecutive inputs.
The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers is separated by one blank and ended by a new-line character. The list of pairs is ended with an end-of-file character. No pair will occur twice.
For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
The output consists of one integer representing the largest number of points that all lie on one line.
1 1 1 2 2 3 3 9 10 10 11
3
题意：问在给出的点中，最多有多少点在一条直线上。
Ps：一定要注意输入，真是坑哭了。UVA的格式错误返回的是WA saddd
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
using namespace std;
const int inf=0x3f3f3f3f;
struct node
{
int x,y;
}q[1010];
char str[1010];
int main()
{
int T,cnt,i,j,k;
int sum,res;
scanf("%d",&T);
getchar();
gets(str);
while(T--){
cnt=0;
res=-inf;
while(gets(str)){
if(str[0]=='\0') break;
sscanf(str,"%d %d",&q[cnt].x,&q[cnt].y);
cnt++;
}
for(i=0;i<cnt;i++)
for(j=i+1;j<cnt;j++){
sum=2;
for(k=j+1;k<cnt;k++)
if((q[j].y-q[i].y)*(q[k].x-q[j].x)==(q[j].x-q[i].x)*(q[k].y-q[j].y))
sum++;
res=max(res,sum);
}
printf("%d\n",res);
if(T)
printf("\n");
}
return 0;
}
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原文地址：http://blog.csdn.net/u013486414/article/details/43561837
