qq_51737737的博客

倍杀测量者//#pragma GCC optimize(2)//#pragma GCC optimize(3,"Ofast","inline")#include<bits/stdc++.h>#define int long long#define fi first#define se second#define pb push_back#define pii pair<int,int>#define IOS ios::sync_with_stdio(false);c

小 K 的农场//#pragma GCC optimize(2)//#pragma GCC optimize(3,"Ofast","inline")#include<bits/stdc++.h>#define int long long#define fi first#define se second#define pb push_back#define pii pair<int,int>#define IOS ios::sync_with_stdio(false)

Returning Home题意：给出一个 n * n 的二维平面，现在要从点 ( sx , sy ) 到达点 ( fx , fy )，在平面上有 m 个传送门，共有两次操作：1，向上、下、左、右移动一个单位，代价为一2，可以瞬间移动到同行或同列的传送门，没有代价。求起点到终点的最小代价。思路：分别按x，y排序，之后建边，跑dij//#pragma GCC optimize(2)//#pragma GCC optimize(3,"Ofast","inline")#include<bit