【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[],g[];

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

int main()
{

        g[] = ;
        for (i = ; i <= ; i++) g[i] = (g[i-] << ) + ;
        f[] = ;
        for (i = ; i <= ; i++)
        {
                f[i] = INF;
                for (j = ; j < i; j++)
                {
                        f[i] = min(f[i],f[j]*+g[i-j]);
                }
        }
        for (i = ; i <= ; i++) printf("%d\n",f[i]);

        return ;

}