题目here
第一道快速幂，同时也是第一道高斯消元。
输入的边的关系矩阵就是系数矩阵co
[co] ^ T * [ans]== (当前0时刻的状态)，[co] ^ T可由矩阵快速幂解得
那么-T时刻的状态便是ans矩阵的值，可由高斯消元解得
判断一下即可
高斯消元中 系数矩阵是a[0...n - 1][0...m - 1] 常数矩阵是a[0...n - 1][m]
返回-1表示无解，等于0有唯一解，大于0表示不确定的变量个数
#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#include <iostream>
#include <algorithm>
#include <bitset>
#include <fstream>
using namespace std;
//LOOP
#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
//OTHER
#define SZ(V) (int)V.size()
#define PB push_back
#define MP make_pair
#define all(x) (x).begin(),(x).end()
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define RS(s) scanf("%s", s)
//OUTPUT
#define WI(n) printf("%d\n", n)
#define WS(n) printf("%s\n", n)
//debug
//#define online_judge
#ifndef online_judge
#define dt(a) << (#a) << "=" << a << " "
#define debugI(a) cout dt(a) << endl
#define debugII(a, b) cout dt(a) dt(b) << endl
#define debugIII(a, b, c) cout dt(a) dt(b) dt(c) << endl
#define debugIV(a, b, c, d) cout dt(a) dt(b) dt(c) dt(d) << endl
#define debugV(a, b, c, d, e) cout dt(a) dt(b) dt(c) dt(d) dt(e) << endl
#else
#define debugI(v)
#define debugII(a, b)
#define debugIII(a, b, c)
#define debugIV(a, b, c, d)
#endif
#define sqr(x) (x) * (x)
typedef long long LL;
typedef unsigned long long ULL;
typedef vector <int> VI;
const double eps = 1e-9;
const int MOD = 1000000007;
const double PI = acos(-1.0);
//const int INF = 0x3f3f3f3f;
const int maxn = 310;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
struct Mat{
int n, m;
bool v[maxn][maxn];
Mat(int n = 0, int m = 0, int zero = 0)
{
this->n = n; this->m = m;
if (zero)
{
REP(i, n) REP(j, m)
v[i][j] = false;
}
}
};
Mat mul(Mat& a, Mat&b)
{
Mat ret(a.n, b.m, 1);
REP(i, a.n)
REP(j, b.m)
REP(k, a.m)
ret.v[i][j] ^= (a.v[i][k] & b.v[k][j]);
return ret;
}
Mat qpow(Mat& a, int b)
{
Mat ret(a.n, a.m);
bool f = 1;
while (b)
{
if (b & 1)
{
if (f)
ret = a, f = 0;
else
ret = mul(ret, a);
}
b >>= 1;
a = mul(a, a);
}
return ret;
}
bool a[maxn][maxn];
int gauss(int N, int M)
{
int r, c, pvt;
bool flag;
for (r = 0, c = 0; r < N && c < M; r++, c++)
{
flag = false;
for (int i = r; i < N; i++)
if (a[i][c])
{
flag = a[pvt = i][c];
break;
}
if (!flag)
{
r--;
continue;
}
if (pvt != r)
for (int j = r; j <= M; j++)
swap(a[r][j], a[pvt][j]);
for (int i = r + 1; i < N; ++i) {
if (a[i][c])
{
a[i][c] = false;
for (int j = c + 1; j <= M; ++j)
if (a[r][j])
a[i][j] = !a[i][j];
}
}
}
for (int i = r; i < N; i++)
if (a[i][M])
return -1;
if (r < M)
return M - r;
for (int i = M - 1; i >= 0; i--)
{
for (int j = i + 1; j < M; j++)
if (a[i][j])
a[i][M] ^= a[j][M];
a[i][M] /= a[i][i];
}
return 0;
}
int main()
{
int n, T, x;
while (~RI(n))
{
Mat co(n, n);
REP(i, n)
REP(j, n)
{
RI(x);
co.v[i][j] = (x == 1 ? true : false);
}
REP(i, n)
{
RI(x);
a[i][n] = (x == 1 ? true : false);
}
RI(T);
co = qpow(co, T);
REP(i, n) REP(j, n) a[i][j] = co.v[i][j];
int ans = gauss(n, n);
if (ans == -1)
puts("none");
else if (ans)
puts("ambiguous");
else
{
REP(i, n)
printf("%d%c", a[i][n], (i == n - 1 ? '\n' : ' '));
}
}
return 0;
}
Graph Automata Player,布布扣,bubuko.com
原文地址：http://blog.csdn.net/colin_27/article/details/38312243
