题意:就是让你求出4个塔的汉诺塔的最小移动步数,(1 <= n <= 12) 那么我们知道3个塔的汉诺塔问题的解为:d[n] = 2*d[n-1] + 1 ,可以解释为把n-1个圆盘移动到一个临时柱子上,然后将1个最大圆盘移动到目标的主子,最后再将n-1个圆盘移动到目标柱子. 为什么是n-1和1的组合呢,因为当你将n-i个圆盘移动到一个临时柱子上的时候,你会发现只靠两个柱子最多能移动1个圆盘.所以这个i=1 如果推到到4个柱子的汉诺塔,f[n] = min(f[n],2*f[n-i]+d[i]…

题目传送门 Strange Towers of Hanoi Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3117   Accepted: 2004 Description Background Charlie Darkbrown sits in another one of those boring Computer Science lessons: At the moment the teacher just exp…

Strange Towers of Hanoi 大体意思是要求\(n\)盘4的的hanoi tower问题. 总所周知,\(n\)盘3塔有递推公式\(d[i]=dp[i-1]*2+1\) 令\(f[i]\)为4塔转移步骤. \(f[i]=min(f[i],f[k]*2+d[i-k])\) 即先以4塔以上面的\(k\),再以3塔移\(i-k\),最后以4塔移动回去. 可以推广到\(n\)盘\(m\)塔 2018.5.26…

Strange Towers of Hanoi Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3784 Accepted: 2376 Description Background  Charlie Darkbrown sits in another one of those boring Computer Science lessons: At the moment the teacher just explains the…

Strange Towers of Hanoi Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 2677 Accepted: 1741 Description Background Charlie Darkbrown sits in another one of those boring Computer Science lessons: At the moment the teacher just explains the…

The teacher points to the blackboard (Fig. 4) and says: "So here is the problem: There are three towers: A, B and C. There are n disks. The number n is constant while working the puzzle. All disks are different in size. The disks are initially stacke…

说是递推,其实也算是个DP吧. 就是4塔的汉诺塔问题. 考虑三塔:先从a挪n-1个到b,把最大的挪到c,然后再把n-1个从b挪到c,所以是 f[i] = 2 * f[i-1] + 1; 那么4塔类似: 先在4塔模式下挪j个到b,然后在三塔模式下挪n-j个到d,然后再在4塔模式下挪j个到d. 故状态转移方程: d[i] = min(2 * d[j] + f[n-j]) ,其中1 <= j < n 初始条件:f[n]与d[1] = 1; 然后就没有然后了. #include <cstdio&…

把M个同样的苹果放在N个同样的盘子里,允许有的盘子空着不放,问共有多少种不同的分法?(用K表示)5,1,1和1,5,1 是同一种分法. Input 第一行是测试数据的数目t(0 <= t <= 20).以下每行均包含二个整数M和N,以空格分开.1<=M,N<=10. Output 对输入的每组数据M和N,用一行输出相应的K. Sample Input 1 7 3 Sample Output 8 题解+代码: 1 /* 2 这道题两种做法,一种dfs,一种递推.这里用的递推法(递推法…

[题目链接] http://poj.org/problem?id=1958 [算法] 先考虑三个塔的情况,g[i]表示在三塔情况下的移动步数,则g[i] = g[i-1] * 2 + 1 再考虑四个塔的情况,f[i]表示在四塔情况下的移动步数,则f[i] = min{2*f[j]+g[i-j]} [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #i…

我对状态空间的理解:https://www.cnblogs.com/AKMer/p/9622590.html 题目传送门:http://poj.org/problem?id=1958 题目要我们求四柱汉诺塔的步数最小值,将盘子数在\(1\)到\(12\)之间的全部求出来. 状态空间即为移动盘子对应的步数. 对于三柱汉诺塔,相信大家都非常熟悉了.我们假设三柱汉诺塔上有\(n\)个盘子,\(f[n]\)表示将\(n\)个盘子移动到另一根柱子上的最小步数,那么显然: \(f[n]=f[n-1]*2+1…