Problem Description The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up o…

1008 Elevator(20 分) The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up o…

[题目描述] 面对蚂蚁们的疯狂进攻,小FF的Tower defense宣告失败……人类被蚂蚁们逼到了Greed Island上的一个海湾.现在,小FF的后方是一望无际的大海,前方是变异了的超级蚂蚁. 小FF还有大好前程,他可不想命丧于此, 于是他派遣手下最后一批改造SCV布置地雷以阻挡蚂蚁们的进攻. 小FF最后一道防线是一条长度为N的战壕, 小FF拥有无数多种地雷,而SCV每次可以在[ L , R ]区间埋放同一种不同于之前已经埋放的地雷. 由于情况已经十万火急,小FF在某些时候可能会询问你在[…

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seco…

畅通工程 http://acm.hdu.edu.cn/showproblem.php?pid=1863 1.Prim算法: Prim算法是由一个点(最初的集合)向外延伸,找到与集合相连权值最小的边, 然后把对应顶点拉到集合里,直到所有的顶点全部在集合里为止. Prim算法的演示如下: http://sjjp.tjuci.edu.cn/sjjg/DataStructure/DS/web/flashhtml/prim.htm #include <stdio.h> ; ; int map[MAXV]…

Friend Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2207    Accepted Submission(s): 1108 Problem Description Friend number are defined recursively as follows. (1) numbers 1 and 2 are friend…

并查集   英文:Disjoint Set,即“不相交集合” 将编号分别为1…N的N个对象划分为不相交集合, 在每个集合中,选择其中某个元素代表所在集合. 常见两种操作: n       合并两个集合 n       查找某元素属于哪个集合 并查集实现的程序代码: int set[MAXN],rank[MAXN]; //set[i]=k表示i的父节点是k,rank[]存储树的深度. int FindSet(int x) { if(set[x]!=x) set[x]=FindSet(set[x])…

A - Space Elevator Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Description The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have…

Space Elevator   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8569   Accepted: 4052 Description The cows are going to space! They plan to achieve orbit by building a sort of space elevator: a giant tower of blocks. They have K (1 <= K…

显然塔的总高度不会超过最大的a[i],而a[i]之前的可以到达的高度 是由a值更小的块组成,所以按照a从小到大的顺序去转移. 然后就是多重背包判断存在性了,几乎和coin那题一样. 数据没coin丧病,倍增可过,单调队列也可,不过还是优化定义最好写. #include<cstdio> #include<iostream> #include<string> #include<cstring> #include<queue> #include<…