POJ 2624 4th Point 计算几何（向量）

题目的链接如下：http://poj.org/problem?id=2624

4th Point
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 5057 Accepted: 1760

Description
Given are the (x,y) coordinates of the endpoints of two adjacent sides of a parallelogram. Find the (x,y) coordinates of the fourth point.

Input
Each line of input contains eight floating point numbers: the (x,y) coordinates of one of the endpoints of the first side followed by the (x,y) coordinates of the other endpoint of the first side, followed by the (x,y) coordinates of one of the endpoints of the second side followed by the (x,y) coordinates of the other endpoint of the second side. All coordinates are in meters, to the nearest mm. All coordinates are between -10000 and +10000.

Output
For each line of input, print the (x,y) coordinates of the fourth point of the parallelogram in meters, to the nearest mm, separated by a single space.

Sample Input

0.000 0.000 0.000 1.000 0.000 1.000 1.000 1.000
1.000 0.000 3.500 3.500 3.500 3.500 0.000 1.000
1.866 0.000 3.127 3.543 3.127 3.543 1.412 3.145

Sample Output

1.000 0.000
-2.500 -2.500
0.151 -0.398

题目的大意如下：给你三个点ABC （我假设输入中给出的相同的那个点是点A)让你求平行四边形的第四个点D（注意这里的三个点不一定就是第二个点和第三个点相同),然后用向量做的话：AB + AC = AD 所以D点的坐标就是B+C-A的坐标，具体的代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#define eps 1e-8struct TPoint{    double x,y;     TPoint operator-(TPoint&a){         TPoint p1;        p1.x=x-a.x;        p1.y=y-a.y;        return p1;    }};using namespace std;int main() {    TPoint ans;    TPoint a[10];    while(~scanf("%lf%lf", &a[0].x, &a[0].y)) {        ans.x = a[0].x, ans.y = a[0].y;        int flag = 1, t = 0;        for(int i = 1; i < 4; i++) {            scanf("%lf%lf", &a[i].x, &a[i].y);            ans.x += a[i].x;            ans.y += a[i].y;            for(int j = 0 ; flag && j < i; j++){                if(a[i].x == a[j].x && a[i].y == a[j].y){                    flag = 0;                    t = i;                }            }        }        ans.x -= 3 * a[t].x;        ans.y -= 3 * a[t].y;        printf("%.3f %.3f\n", ans.x, ans.y);    }    return 0;}