Codeforces Round #528 (Div. 2) C. Connect Three 思维

题解

题目大意 一个格子图 给三个人坐标 问联通所需要开发的最少方块数量

按照纵坐标排序 abc分别为左中右 将b的位置纵向扩展到a和c的高度 然后a和c向b横向汇聚

AC代码

#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;

int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	int ax, ay, bx, by, cx, cy; //按照纵坐标排序 abc分别为左中右
	cin >> ax >> ay >> bx >> by >> cx >> cy;
	int top = max({ ax, bx, cx });
	int down = min({ ax, bx, cx });
	if (ay > by)
		swap(ax, bx), swap(ay, by);
	if (ay > cy)
		swap(ax, cx), swap(ay, cy);
	if (by > cy)
		swap(bx, cx), swap(by, cy);
	vector<pair<int, int>> ans;
	for (int i = down; i <= top; i++) //将b的位置纵向扩展到a和c的高度
		ans.push_back({ i, by });
	for (int i = ay; i < by; i++) //横向想b汇聚
		ans.push_back({ ax, i });
	for (int i = by + 1; i <= cy; i++)
		ans.push_back({ cx, i });
	cout << ans.size() << endl;
	for (int i = 0; i < ans.size(); i++)
		cout << ans[i].first << " " << ans[i].second << endl;

	return 0;
}