Okabe and City 题解: 将行和列也视为一个点. 然后从普通的点走到行/列的点的话,就代表这行/列已经被点亮了. 然后将费用为0的点建上边. 注意讨论(n,m)非亮的情况下. 代码: #include<bits/stdc++.h> using namespace std; #define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w"…

其实,这道题不用long long也能AC. 题意是给你一个矩阵,有一些格子被点亮有一些没有,每一次只能在被点亮的格子上面走. 然后你每一次都可以选择点亮一行或一排(非永久),现在问你最少点多少次可以走到终点? 思路十分好想. 我们把相邻的格子边权设为0,把不相邻但只差一行的格子之间边权设为1.(其他情况都不能直接到达) 然后跑一下SPFA就可以了. 但是需要考虑一个特例:终点有没有被点亮. 如果点了的话那没关系,没点的话得把(n+1,m+1)设为点亮的点.(要不然我们走不到终点不是吗) AC代…

题目链接:http://codeforces.com/contest/821/problem/D 题意:n*m地图,有k个位置是点亮的,有4个移动方向,每次可以移动到相邻的点亮位置,每次站在初始被点亮某个位置,暂时使某行或该某列全部点亮,花费为1,下一次使用时,上一次暂时点亮被熄灭 题解:显然只要知道如果两点相邻直接相连然后就是极限情况,什么情况之下是两点是连不到一起的.如果x轴与y轴的相差大于2是连不到一起的. 具体画一下图就知道了.然后就跑一遍dij+优先队列就行了或者spfa也行. #in…

Link 题意:给出$n*m$大小的地图,已有$k$盏灯亮,人从左上角出发,右下角结束,期间必须走路灯点亮的地方,他可以在任意时刻消耗一枚硬币点亮一行或一列灯,他最多同时点亮一行或一列灯,要想点亮别的行列时,原先灯的状态将还原. 思路:看似很繁琐的题目,其实重点在于每次只能进行一次操作,那么只需要边走边考虑下一步到达的点即可.由于可任选行列,那么意味着下一个点只要和当前位置x和y坐标之差都不超过2,那么都能到达,在此情况上,坐标差之和为1说明相邻不需要消耗硬币,其余情况均消耗一枚硬币.跑个最短路…

E. Okabe and El Psy Kongroo     Okabe likes to take walks but knows that spies from the Organization could be anywhere; that's why he wants to know how many different walks he can take in his city safely. Okabe's city can be represented as all points…

E. Okabe and El Psy Kongroo time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Okabe likes to take walks but knows that spies from the Organization could be anywhere; that's why he wants to k…

E. Okabe and El Psy Kongroo time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Okabe likes to take walks but knows that spies from the Organization could be anywhere; that's why he wants to k…

E. Okabe and El Psy Kongroo   Okabe likes to take walks but knows that spies from the Organization could be anywhere; that's why he wants to know how many different walks he can take in his city safely. Okabe's city can be represented as all points (…

/****************************************************************************************************************** 因为发现不敲题会变蠢...所以没事还是做两道题吧.... 很久没做过题然后发现C和E全都是因为没用LL给卡住了......  我真的是太蠢了 ***************************************************************…

本来打算划划水洗洗睡了,突然听到这次的主人公是冈部伦太郎 石头门(<steins;gate>)主题的比赛,岂有不打之理! 石头门真的很棒啊!人设也好剧情也赞曲子也特别好听. 推荐http://music.163.com/#/m/song?id=26259014&userid=115264555 (强行跑题) Okabe and Future Gadget Laboratory O(n^4)暴力妥妥的 #include<iostream> #include<algori…