【POJ 1958】 Strange Towers of Hanoi

【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>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h> 
using namespace std;
const int INF = 2e9;

int i,j;
int f[13],g[13];

template <typename T> inline void read(T &x)
{
    int f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) { if (c == '-') f = -f; }
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;
}
template <typename T> inline void write(T x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x/10);
    putchar(x%10+'0');
}
template <typename T> inline void writeln(T x)
{
    write(x);
    puts("");
}

int main() 
{
        
        g[1] = 1;
        for (i = 2; i <= 12; i++) g[i] = (g[i-1] << 1) + 1;
        f[1] = 1;
        for (i = 2; i <= 12; i++)
        {
                f[i] = INF;
                for (j = 1; j < i; j++)
                {
                        f[i] = min(f[i],f[j]*2+g[i-j]);
                }
        }
        for (i = 1; i <= 12; i++) printf("%d\n",f[i]);
        
        return 0;
    
}