考虑到每次与主人相遇之前最多只去一个景点,很容易转化为匹配问题 由于数据很小,我们不妨枚举每个相遇点间隙和每个景点,判断是否来得及,如果来得及就连边 沙雕题搞了二十来分钟,我是憨憨 #include <bits/stdc++.h> using namespace std; #define N 505 int n,m,p,cx[N],cy[N],vis[N],ax[N],ay[N],bx[N],by[N]; std::vector<int> e[N]; int dfs(int u,i…
在开始看我画小狗之前,咱们先来看道很简单的题目: 下面程序的输出是什么? Dog myDog = new Dog("旺财"); changeName(myDog); System.out.println(myDog.getName()); public void changeName(dog) { dog.setName("小强"); } 如果你的回答是“小强”,好,恭喜你答对了.下面我们改一下代码: Dog myDog = new Dog("旺财&quo…
Tempter of the Bone Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggi…
Problem Description The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried despe…
