Codeforces 514B Han Solo and Lazer Gun【思维】

There are n Imperial stormtroopers on the field. The battle field is a plane with Cartesian coordinate system. Each stormtrooper is associated with his coordinates(x, y) on this plane.

Han Solo has the newest duplex lazer gun to fight these stormtroopers. It is situated at the point(x₀, y₀). In one shot it can can destroy all the stormtroopers, situated on some line that crosses point(x₀, y₀).

Your task is to determine what minimum number of shots Han Solo needs to defeat all the stormtroopers.

The gun is the newest invention, it shoots very quickly and even after a very large number of shots the stormtroopers don't have enough time to realize what's happening and change their location.

The first line contains three integers n,x₀ и y₀ (1 ≤ n ≤ 1000, - 10⁴ ≤ x₀, y₀ ≤ 10⁴) — the number of stormtroopers on the battle field and the coordinates of your gun.

Next n lines contain two integers each x_i, y_i ( - 10⁴ ≤ x_i, y_i ≤ 10⁴) — the coordinates of the stormtroopers on the battlefield. It is guaranteed that no stormtrooper stands at the same point with the gun. Multiple stormtroopers can stand at the same point.

PRint a single integer — the minimum number of shots Han Solo needs to destroy all the stormtroopers.

4 0 01 12 22 0-1 -1Output
2Input
2 1 21 11 0Output
1Note
Explanation to the first and second samples from the statement, respectively: 
题目大意：
一共有N个目标，给出N个目标点的坐标，并且给出机枪的目标。
这个机枪非常牛X，其开一枪，前后的人都会死。
问最少开多少枪，能够将所有目标都打到。
思路：
问题寻找共性。
对于一条路径上的所有目标都有一个唯一的共性，那就是斜率。
那么我们对问题的斜率进行处理即可。任务目标就是统计斜率的个数。
K=（y-y0）/（x-x0）；
注意这个K需要用double类型。
过程用map维护一下第一次出现即可。
Ac代码：
#include<stdio.h>#include<string.h>#include<map>using namespace std;int main(){    int n;    int x,y;    while(~scanf("%d%d%d",&n,&x,&y))    {        int flag1=0;        int flag2=0;        int output=0;        map<double ,int >s;        for(int i=0;i<n;i++)        {            int xx,yy;            scanf("%d%d",&xx,&yy);            if(xx==x)            {                flag1=1;                continue;            }            if(yy==y)            {                flag2=1;                continue;            }            double  k=(yy-y)*1.0/(xx-x)*1.0;            if(s[k]==0)            {                s[k]=1;                output++;            }        }        printf("%d/n",output+flag1+flag2);    }}