Water pipe Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 2265 Accepted: 602 Description The Eastowner city is perpetually haunted with water supply shortages, so in order to remedy this problem a new water-pipe has been built. Builders s…

POJ 1129 Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14191   Accepted: 7229 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receive…

[题目描述] An addition chain for n is an integer sequence with the following four properties: a0 = 1 am = n a0 < a1 < a2 < ... < am-1 < am For each k (1<=k<=m) there exist two (not necessarily different) integers i and j (0<=i, j<=k…

题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=1085 题目大意: 在一个5×5的棋盘上有12个白色的骑士和12个黑色的骑士, 且有一个空位.在任何时候一个骑士都能按照骑士的走法(它可以走到和它横坐标相差为1,纵坐标相差为2或者横坐标相差为2,纵坐标相差为1的格子)移动到空位上. 给定一个初始的棋盘,怎样才能经过移动变成如下目标棋盘: 为了体现出骑士精神,他们必须以最少的步数完成任务. 思路: 迭代加深搜索+A*剪枝即可. 每次枚举…

题目链接:点击打开链接 题目描写叙述: 在古埃及.人们使用单位分数的和(形如1/a的, a是自然数)表示一切有理数.如:2/3=1/2+1/6,但不同意2/3=1/3+1/3,由于加数中有同样的.对于一个分数a/b,表示方法有非常多种.可是哪种最好呢?首先.加数少的比加数多的好,其次.加数个数同样的,最小的分数越大越好. 如:19/45=1/3 + 1/12 + 1/180 19/45=1/3 + 1/15 + 1/45 19/45=1/3 + 1/18 + 1/30, 19/45=1/4 +…

Channel Allocation Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14601   Accepted: 7427 Description When a radio station is broadcasting over a very large area, repeaters are used to retransmit the signal so that every receiver has a s…

1085: [SCOI2005]骑士精神 Time Limit: 10 Sec  Memory Limit: 162 MBSubmit: 1800  Solved: 984[Submit][Status][Discuss] Description 在一个5×5的棋盘上有12个白色的骑士和12个黑色的骑士, 且有一个空位.在任何时候一个骑士都能按照骑士的走法(它可以走到和它横坐标相差为1,纵坐标相差为2或者横坐标相差为2,纵坐标相差为1的格子)移动到空位上. 给定一个初始的棋盘,怎样才能经过移动变…

codevs 2541 幂运算  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 钻石 Diamond 题目描述 Description 从m开始,我们只需要6次运算就可以计算出m31: m2=m×m,m4=m2×m2,m8=m4×m4,m16=m8×m8,m32=m16×m16,m31=m32÷m. 请你找出从m开始,计算mn的最少运算次数.在运算的每一步,都应该是m的正整数次方,换句话说,类似m-3是不允许出现的. 输入描述 Input Description 输入为一…

题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=1560 BFS题解:http://www.cnblogs.com/crazyapple/p/3218107.html 构造一个串,使得它包含所有的给定DNA序列,并且要求长度最短.采用dfs策略,对于每个串定义1个指针,当全部指针为串尾时判断构造的长度,由于状态空间过大,但是又存在冗余搜索,可以用迭代加深将状态减少.最长待构造长度 + 当前长度 < 迭代的最大深度则直接return,大大减少了状态数.…

Description An addition chain for n is an integer sequence  with the following four properties: a0 = 1 am = n a0<a1<a2<...<am-1<am For each k ( ) there exist two (not neccessarily different) integers i and j ( ) with ak =ai +aj You are give…