Codeforces Round #388 (Div. 2)B Parallelogram is Back

题目大意：

      给出平行四边形的3个点，求第4个点的所有可能性。

题目解法：

      3个点3条边，任意取一条作为平行四边形的边剩下一个点作为平行边的端点，就可以找出剩下一个点，为了防止重复，用map记录。

代码：

#include "iostream"#include "cstdio"#include "math.h"#include "algorithm"#include "string"#include "string.h"#include "vector"#include "map"#include "queue"#include "bitset"using namespace std;struct Node {int x, y;Node() {}Node(int dx, int dy) {x = dx;y = dy;}}node[4];vector<Node>ans;map<pair<int, int>, int>mp;int main() {for (int i = 0;i < 3;i++) {scanf("%d %d", &node[i].x, &node[i].y);}for (int i = 0;i < 3;i++) {for (int j = i + 1;j < 3;j++) {int next = 3 - i - j;int x1 = node[next].x + (node[i].x - node[j].x);int x2 = node[next].x - (node[i].x - node[j].x);int y1 = node[next].y + (node[i].y - node[j].y);int y2 = node[next].y - (node[i].y - node[j].y);if (mp.count(make_pair(x1,y1))==0) {mp[make_pair(x1, y1)] = 1;ans.push_back(Node(x1, y1));}if (mp.count(make_pair(x2, y2)) == 0) {mp[make_pair(x2, y2)] = 1;ans.push_back(Node(x2, y2));}}}printf("%d\n", ans.size());for (int i = 0;i < ans.size();i++) {printf("%d %d\n", ans[i].x, ans[i].y);}return 0;}