Hanoi
11.3
类比只有三个柱子的汉诺塔, 设f[i][j]为有i个盘子j个柱子时的最少步数. 那么肯定是把一些上面盘子移动到某根不是j的柱子上, 然后把剩下的盘子移动到j, 然后再把上面的盘子移动到j. 于是就有递推式f[i][j] = min{f[k][j] * 2 + f[i - k][j - 1]}.
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define LL long long
#define N 66
using namespace std;
int n, m;
LL f[N][N];
int main() {
freopen( "hanoi.in", "r", stdin);
freopen( "hanoi.out", "w", stdout);
memset(f, -1, sizeof(f));
for(int i=1; i<65; ++i) f[1][i] = 1;
for(int i=2; i<64; ++i) {
for(int j=3; j<=i+1; ++j)
for(int k=1; k<i; ++k) {
if(f[k][j] == -1 || f[i-k][j-1] == -1) continue;
LL cc = (f[k][j] << 1) + f[i-k][j-1];
if(f[i][j] == -1 || f[i][j] > cc) f[i][j] = cc;
}
for(int j=i+2; j<65; ++j) f[i][j] = f[i][j-1];
}
cin >> n >> m;
printf("%lld\n", f[n][m]);
return 0;
}