hdu 1548 A strange lift 解题报告】的更多相关文章

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 题目意思:给出 n 个 floor 你,每个floor 有一个数k,按下它可以到达 floor + k 或者 floor - k的位置.问从floor A 到 floor  B 最少的按lift 次数是多少. hdu 真是!!!!! queue<node>  q 写在main 外就 wa了!!! = = 汗!!! 还专门瞪大双眼对照别人AC的代码,看了一遍又一遍,以为色盲了= =,可恶HDU…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1548 A strange lift Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number $K_i(0 \leq K_i \leq N)$ on every floor.The lift have just two buttons: up…
A strange lift http://acm.hdu.edu.cn/showproblem.php?pid=1548 Problem Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if yo…
题目链接: 传送门 A strange lift Time Limit: 1000MS     Memory Limit: 32768 K Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21754    Accepted Submission(s): 7969 Problem Description There is a strange li…
A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11341    Accepted Submission(s): 4289 Problem Description There is a strange lift.The lift can stop can at every floor as you want…
A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8550    Accepted Submission(s): 3241 Problem Description There is a strange lift.The lift can stop can at every floor as you want,…
Problem Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP"…
题目链接 Problem Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP…
A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26943    Accepted Submission(s): 9699 Problem Description There is a strange lift.The lift can stop can at every floor as you want,…
A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33697    Accepted Submission(s): 12038 Problem Description There is a strange lift.The lift can stop can at every floor as you want…
A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 18723    Accepted Submission(s): 6926 Problem Description There is a strange lift.The lift can stop can at every floor as you want,…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 题目大意:升降电梯,先给出n层楼,然后给出起始的位置,即使输出从A楼道B楼的最短时间. 注意的几点 (1)每次按一下,只能表示上或者是下,然后根据输入的看是上几层或者是下几层. (2)注意不能到底不存在的楼层. 详见代码. #include <iostream> #include <cstdio> using namespace std; ; ][],node[],vis[],M…
题目大意: 电梯有两个选项向上或向下,每层楼有一个参数ki,代表电梯可以再该楼层的基础上向上或向下移动ki层,限制条件是向上不能超过楼层总数n,向下不能少于一.输入总层数n和当前所在层数以及目标层数,然后是n个数分别代表第i层的移动范围.输出最少移动次数,若不可达,输出-1. 解题思路: 1.用Dijkstra算法,首先构建邻接矩阵,注意在构造时,要考虑i-k[i]<1和i+k[i]>n,i代表当前所在层. #include<string.h> #include<stdio.…
题意: 有一座电梯,其中楼层从1-n,每层都有一个数字k,当处于某一层时,只能往上走k层,或者下走k层.楼主在a层,问是否能到达第b层? 思路: 在起点时只能往上走和往下走两个选择,之后的每层都是这样,那么就类似于二叉树.每个节点就是对应的层,因为有可能碰到循环的层,比如1跳到3,3跳回1,这样使得无限循环,所以加个vis数组标记是否遍历过即可. #include <iostream> #include <cmath> #include <cstring> #inclu…
( ̄▽ ̄)" //dijkstra算法, //只是有效边(即能从i楼到j楼)的边权都为1(代表次数1): //关于能否到达目标楼层b,只需判断最终lowtime[b]是否等于INF即可. #include<iostream> #include<cstdio> using namespace std; const int INF=10e7; ; int k,minn; int K[MAXN]; int cost[MAXN][MAXN]; int lowtime[MAXN];…
https://vjudge.net/problem/HDU-1548 题意: 电梯每层有一个不同的数字,例如第n层有个数字k,那么这一层只能上k层或下k层,但是不能低于一层或高于n层,给定起点与终点,要求出最少要按几次键. 思路: 可以用BFS,也可以用迪杰斯特拉算法. #include<iostream> #include<cstring> #include<algorithm> #include<vector> using namespace std;…
题意:给出一个电梯,给出它的层数f,给出起点s,终点g,以及在每一层能够上或者下w[i]层,问至少需要按多少次按钮到达终点. 和POJ catch that cow一样,直接用了那一题的代码,发现一直wa, 后来才发现,POJ catch that cow是单组输入的,所以每次调用的时候不用清空队列,而这一题一次输入有多组数据--- 用这个清空队列 while(!q.empty()) q.pop(); #include<iostream> #include<cstdio> #inc…
http://acm.hdu.edu.cn/showproblem.php?pid=1548 A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11341    Accepted Submission(s): 4289 Problem Description There is a strange lift.T…
Problem Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP"…
A strange lift Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29442    Accepted Submission(s): 10641 Problem Description There is a strange lift.The lift can stop can at every floor as you want…
Strange fuction Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5933 Accepted Submission(s): 4194 Problem Description Now, here is a fuction: F(x) = 6 * x^7+8*x^6+7*x^3+5*x^2-y*x (0 <= x <=100) C…
1.题目描述: 1175. Strange Sequence Time limit: 1.0 secondMemory limit: 2 MB You have been asked to discover some important properties of one strange sequences set. Each sequence of the parameterized set is given by a recurrent formula: Xn+1 = F(Xn-1, Xn)…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1556 题目意思:有 n 个气球从左到右排成一排,编号依次为1,2,3,...,n.给出 n 对 a, b,表示编号为 a ~b 的气球被涂过一次色.n 次之后,问所有气球被上色的次数分别是多少. 典型的树状数组题!这种是:每次修改一个区间,问的是最后每个点被修改的次数.要注意update() 函数对 a 处理 的时候,向上修改的过程中把不必要的区间都加了一遍.例如对a = 2时,update修改了c…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1160 题目意思:给出一堆老鼠,假设有 n 只(输入n条信息后Ctrl+Z).每只老鼠有对应的weight 和 speed.现在需要从这 n 只老鼠的序列中,找出最长的一条序列,满足老鼠的weight严格递增,speed严格递减. 我们可以用一个结构体来保存老鼠的信息,包括weight, speed 以及 id(这个 id 是未排序之前的,为了输出最后信息).那么首先对 weight 进行递增排序,如…
题目链接:http://code.hdu.edu.cn/showproblem.php?pid=1879 这条题目我的做法与解决Constructing Roads的解法是相同的. 0 表示没有连通:1代表已连通.在已连通村庄的道路的基础上,找到扩展出来的最小生成树. #include <iostream> #include <algorithm> using namespace std; + ; int rep[maxe], vis[maxe]; struct sets { in…
题目链接:http://code.hdu.edu.cn/showproblem.php?pid=1233 并查集的运用, 实质就是求最小生成树.先对所有的村庄距离从小到大排序,然后判断村庄之间是否属于同一集合,不是则将距离相加.属于同一集合,说明村庄连成了环,就不符合树的定义了.这样扫描下来,就求得最小生成树了. 要特别注意的是,数组越界问题.这个得益于Dwylkz的指点.由于村庄最多假设为100,为了防止边数越界,数组应该开到100 * 100 (10000)左右. #include <ios…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1213 有关系(直接或间接均可)的人就坐在一张桌子,我们要统计的是最少需要的桌子数. 并查集的入门题,什么优化都无用到就可以直接过.实质上就是统计总共有多少个不相交的集合.比较可恶的是,题目中 There will be a blank line between two cases 是骗人的!!! #include <iostream> using namespace std; + ; int p[…
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1272 第二条并查集,和畅通工程的解法类似.判断小希的迷宫不符合条件,即有回路.我的做法是,在合并两个集合的时候,当fx = fy,即有共同祖先的时候,说明就有回路. 这题有三点要注意:1.格式问题.题目说的“每两组数据之间有一个空行.”是会PE的!!实际上输出Yes或No之后加多个\n即可,不需要再画蛇添足再输多一个换行.  2. 当整数对列表只有0 0 时,要输出Yes   3.当不相交的集合个数…