链接 Codeforces 677D Vanya and Treasure 题意 n*m中有p个type,经过了任意一个 type=i 的各自才能打开 type=i+1 的钥匙,最初有type=1的钥匙, 问拿到type=p的钥匙最少需要走多少步 思路 第一想法就是按type来递推, 将type相同的存到一起,dp[i][j]=min(dp[i][j], dp[k][l]+distance([i][j], [k][l])),其中 a[i][j] = a[k][l]+1. 但这样type相同的个数…

677D. Vanya and Treasure 题意: 给定一张n*m的图,图上每个点标有1~p的值,你初始在(1,1)点,你必须按照V:1,2,3...p的顺序走图上的点,问你如何走时间最少. 思路: 我一开始想的思路感觉很巧妙,但是TLE了.就是把不同值的点放在不同的vector中,然后类似dp的从2更新最小距离到p.因为我这是暴力枚举点的,复杂度不对.后来发现这个思路还需要优化一下,就是把同一行的点放在一起,for一遍这一行属于V的点,就可以更新本行的信息了,再向下把列中属于v+1的更新…

题目链接:http://codeforces.com/problemset/problem/677/D 题意: 有 $n \times m$ 的网格,每个网格上有一个棋子,棋子种类为 $t[i][j]$,棋子的种类数为 $p$. 现在出发点为 $(1,1)$,必须按照种类 $1 \sim p$ 进行移动,即从种类 $x$ 的棋子出发,下一个目标必须是 $x+1$ 才行,直到走到种类为 $p$ 的棋子就终止.求最短路径. 题解: 我们先把棋子按照种类分组,分成 $p$ 组. $dp[i][j]$…

$dp$,树状数组. 很明显这是一个$DAG$上的$dp$,由于边太多,暴力$dp$会超时,需要优化. 例如计算$dp[x][y]$,可以将区域分成四块,$dp[x][y]$取四块中的最小值,每一块用一个二维树状数组维护最小值即可. 每次扩展一层需要一个新的树状数组,因为每次初始化树状数组会超时,所以可以额外开一个数组记录一下每一个点是第几次更新的. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<…

D. Vanya and Treasure 题目连接: http://www.codeforces.com/contest/677/problem/D Description Vanya is in the palace that can be represented as a grid n × m. Each room contains a single chest, an the room located in the i-th row and j-th columns contains t…

题目链接: D. Vanya and Treasure time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output Vanya is in the palace that can be represented as a grid n × m. Each room contains a single chest, an the room…

CF677D Vanya and Treasure 有一个 \(n\times m\) 的矩阵 \(a(1\le a_{i,j}\le p)\),求从起点 \((1,1)\) 出发依次遍历值为 \(1\to p\) 的矩阵单元的最短路径曼哈顿距离.保证满足 \(a_{i,j}=p\) 的 \((i,j)\) 唯一. 数据范围:\(1\le n,m\le 300\),\(1\le p\le n\cdot m\). 先记录 \(\tt vector\) 数组 \(w\),\(w_t\) 表示 \(a…

Open the Lock Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3847    Accepted Submission(s): 1661 Problem Description Now an emergent task for you is to open a password lock. The password is co…

题目链接:codeforces 492e vanya and field 留个扩展gcd求逆元的板子. 设i,j为每颗苹果树的位置,因为gcd(n,dx) = 1,gcd(n,dy) = 1,所以当走了n步后,x从0~n-1,y从0~n-1都访问过,但x,y不相同. 所以,x肯定要经过0点,所以我只需要求y点就可以了. i,j为每颗苹果树的位置,设在经过了a步后,i到达了0,j到达了M. 则有 1----------------------(i + b * dx) % n = 0 2------…

Codeforces 677D 传送门:https://codeforces.com/contest/677/problem/D 题意: 给你一个n*m的方格图,每个点有一个权值val,现在要求你从坐标(1,1)开始走,要求你从权值为1的点,走到权值为2的点,依次类推,最终走到权值为p的点的最短路径是多少 题解: 分层图dp \[ dp[i][j]表示到达点(i,j)所需要的最短距离是多少\\ dis维护一个纵列上的距离\\ vis维护一个当前走到的位置\\ 转移:dp[r][c] = min(…

time limit per test1.5 seconds memory limit per test256 megabytes inputstandard input outputstandard output Vanya is in the palace that can be represented as a grid n × m. Each room contains a single chest, an the room located in the i-th row and j-t…

A. Vanya and Table Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/552/problem/A Description Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom…

题目大意: 给你一个n × m 的图,有p种宝箱, 每个点上有一个种类为a[ i ][ j ]的宝箱,a[ i ][ j ] 的宝箱里有 a[ i ][ j ] + 1的钥匙,第一种宝箱是没有锁的, 第p类宝箱只有一个且里面由宝藏,你现在在(1 ,1)问你最少需要多少步才能拿到宝藏. (n, m <= 300) 思路:这题真的好恶心啊...  我们考虑p类宝箱只能从p - 1类转移过来, 这样我们就能从第一类宝箱开始往后递推, 但是最坏的情况, p 类 和p - 1类,都有45000 个点, 那…

题目链接: http://codeforces.com/contest/677/problem/D 题意: 让你求最短的从start->...->1->...->2->...->3->...->...->p的最短路径. 题解: 这题dp的阶段性还是很明显的,相同的值得方格为同一个阶段,然后求从阶段1->2->3...->p的阶段图最短路. 初始化所有a[x][y]==1的格子为起始点到(x,y)坐标的距离. 方程式为dp[x1][y1…

B. Biridian Forest Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/329/problem/B Description You're a mikemon breeder currently in the middle of your journey to become a mikemon master. Your current obstacle is go through t…

http://codeforces.com/contest/677/problem/D 建颗新树,节点元素包含r.c.dis,第i层包含拥有编号为i的钥匙的所有节点.用i-1层更新i层,逐层更新到底层. 不使用就会超时的优化:用i-1层更新时不是所有节点都有必要用到,我们对i-1层排序,取前600节点更新下层. public class Main { private static final int c = 330,INF=Integer.MAX_VALUE/2,maxn=c*c*c+100,m…

解题思路: 1.对物品i bfs,更新每个小镇j获得每个物品i的最短距离. 2.时间复杂度o(n*k),满足2s的要求. 代码: #include <iostream> #include <queue> #include <list> #include <algorithm> #include <stdio.h> #include <string.h> using namespace std; typedef long long ll…

Vanya and Brackets Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Description Vanya is doing his maths homework. He has an expression of form , where x1, x2, ..., xn are digits from 1 to 9, and sign represents either a p…

Vanya and Scales Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 552C Description Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some…

Hidden Code 题目连接: http://codeforces.com/gym/100015/attachments Description It's time to put your hacking skills to the test! You've been called upon to help crack enemy codes in the current war on... something or another. Anyway, the point is that yo…

E. Two Labyrinths Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100187/problem/E Description A labyrinth is the rectangular grid, each of the cells of which is either free or wall, and it's possible to move only between free…

题目链接: http://www.lydsy.com/JudgeOnline/problem.php?id=1295 题解: 对每个点暴力跑一遍bfs,看能够到达的最远位置,这里如果有障碍物则距离为1,如果没有障碍物,则距离为0,用bfs跑距离<=t的所有点并更新答案. 代码: #include<iostream> #include<cstring> #include<cstdio> #include<utility> #include<queu…

A. ArielTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100685/problem/A Description King Triton really likes watching sport competitions on TV. But much more Triton likes watching live competitions. So Triton decides to set up…

G. #TheDress Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100637/problem/G Description After landing on planet i1c5l people noticed that blue and black clothes are quite popular among the locals. Each aboriginal has at least…

Couple Cover Time Limit: 3000MS   Memory Limit: 524288KB   64bit IO Format: %I64d & %I64u Description 方宝宝有n个篮球,每个篮球上写有一个值ai.他第一次从n个篮球中取出1个,不放回.第二次再在剩余的篮球中取出一个. (每个球被取概率相同).如果这两个球上的值的乘积大于等于p,他会变得高兴,然后请大家吃饭.否则方宝宝会不高兴, 然后暴食暴饮变得更胖. 当然为了方宝宝的健康(被请吃饭),我想取一个…

D. Jerry's Protest time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Andrew and Jerry are playing a game with Harry as the scorekeeper. The game consists of three rounds. In each round, Andr…

https://vjudge.net/problem/CodeForces-516B 题意 在一个n*m图中放1*2或者2*1的长方形,问是否存在唯一的方法填满图中的‘.’ 分析 如果要有唯一的方案,那么必定存在度为一的点,因为只有这样,把这一格以及它相邻的涂掉的方案才唯一,然后可能产生新的度为一的可行点,不断更新,bfs寻找这样的点.最后检测一遍是否还有‘.'存在即可. #include<iostream> #include<cmath> #include<cstring&…

C. Vanya and Label time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output While walking down the street Vanya saw a label "Hide&Seek". Because he is a programmer, he used & as a bitwi…

题意:给定4个数,a,b,c,d,分别代表空杯子容积为a,b,一个盛满水的杯子容积为c,让你不断倒水,找一个dd,是不是存在某个时刻, 某个杯子里的水dd,和d相同,或者无限接近.让求最少的倒水量和dd(可能和d相同). 析:首先由于没有刻度,如果用数学方法计算,不好算,样例还好算一点,我们观察那个a,b,c都不大于200,挺小的,适合暴力求解. 就是把所有情况都倒一次,倒水就两种倒法,要么把一个杯子倒满,要么就是这个杯子空了,由于水量是固定的,那么确定两个杯子的水量, 那么第三个也就确定了,所…

Description Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon's home. But he can't bring his precious with him. As you know, uncle Ve…