思路简单,把状态看成一个长十五的01串,总状态为(2^15),以判断每个小球可否转移,转移的费时接近(3*11*4)大概最坏七八百万算法;总共输入为1-15打表可过;
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 16;
vector<int> a[maxn][7];
int deg[maxn],n;
const int b[11][6]={
{1,2,4,7,11},
{3,5,8,12},
{6,9,13},
{10,14},
{2,3},
{4,5,6},
{7,8,9,10},
{1,3,6,10,15},
{2,5,9,14},
{4,8,13}......
阅读全文