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题目大意
给一个4*n的矩阵,矩阵看成无向图,求哈密顿回路方案数.
题解
看了题解才懂.不然身为蒟蒻的我怎么可能推出递推式23333
附网址http://blog.csdn.net/jiangshibiao/article/details/21446033
代码和上面的有所不同.
代码
/* TASK:vans LANG:C++ */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; struct Bigint { static const int BASE = 10000; int s[200], len; Bigint() { memset(s, 0, sizeof(s)); len = 1; } Bigint operator + (const Bigint &b) { Bigint c; c.len = max(len, b.len); for (int i = 0; i < c.len; ++i) { c.s[i] += s[i] + b.s[i]; c.s[i + 1] = c.s[i] / BASE; c.s[i] %= BASE; } if (c.s[c.len]) c.len++; return c; } }; int n; Bigint f[1005], g[1005], s[1005]; int main() { freopen("vans.in", "r", stdin); freopen("vans.out", "w", stdout); scanf("%d", &n); f[1].s[0] = 0; f[2].s[0] = 2; g[1].s[0] = 2; s[1].s[0] = 0; g[2].s[0] = 2; s[2].s[0] = 2; for (int i = 3; i <= n; ++i) { f[i] = f[i - 1] + g[i - 1]; s[i] = f[i - 2] + f[i - 2] + s[i - 1]; g[i] = f[i - 1] + f[i - 1] + g[i - 2] + s[i]; } printf("%d", f[n].s[f[n].len - 1]); for (int i = f[n].len - 2; i >= 0; --i) printf("%04d", f[n].s[i]); printf("\n"); return 0; }
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原文地址:http://www.cnblogs.com/albert7xie/p/5646571.html