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;
}