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

題目大意：

      給出平行四邊形的3個點，求第4個點的所有可能性。

題目解法：

      3個點3條邊，任意取一條作爲平行四邊形的邊剩下一個點作爲平行邊的端點，就可以找出剩下一個點，爲了防止重複，用map記錄。

代碼：

#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;
}