链接 把迭代加深理解错了 自己写了半天也没写对 所谓迭代加深,就是在深度无上限的情况下,先预估一个深度(尽量小)进行搜索,如果没有找到解,再逐步放大深度搜索.这种方法虽然会导致重复的遍历 某些结点,但是由于搜索的复杂度是呈指数级别增加的,所以对于下一层搜索,前面的工作可以忽略不计,因而不会导致时间上的亏空. IDA*就是一个加了层数限制depth的DFS,超过了限制就不在搜索下去,如果在当前层数没有搜到目标状态,就加大层数限制depth,这里还只是一个IDA算法,并不是A*的.当然我们可以用A*…

Description Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications: x2 = x × x, x3 = x2 × x, x4 = x3 × x, …, x31 = x30 × x. The operation of squaring can be appreciably shorten the sequence of multiplications.…

An addition chain for n is an integer sequence <a0, a1,a2,...,am=""> 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) int…

题目链接:http://bailian.openjudge.cn/practice/2248 题解: 迭代加深DFS. DFS思路:从目前 $x[1 \sim p]$ 中选取两个,作为一个新的值尝试放入 $x[p+1]$. 迭代加深思路:设定一个深度限制,一旦到达这个界限,即继续往下搜索:该深度限制从 $1$ 开始,每次自加 $1$.这么做的好处是,正好也符合题目要求的最短的数组长度. AC代码: #include<bits/stdc++.h> using namespace std; ];…

Description Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications: x2 = x × x, x3 = x2 × x, x4 = x3 × x, -, x31 = x30 × x. The operation of squaring can be appreciably shorten the sequence of multiplications.…

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…

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…

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…

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…