ZOJ 3497 Mistwald 矩阵

利用可达矩阵的幂来判断是否可达

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const LL MOD = LINF;
const double DINF = 1e60;
const int maxn = 30;

struct Matrix {
    int n, m;
    LL data[maxn][maxn];
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }

    void print() {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                cout << data[i][j] << " ";
            }
            cout << endl;
        }
    }
};
 
Matrix operator * (Matrix a, Matrix b) {
    Matrix ret(a.n, b.m);
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= b.m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
                ret.data[i][j] %= MOD;
            }
        }
    }
    return ret;
}
 
Matrix operator + (Matrix a, Matrix b) {
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= a.m; j++) {
            a.data[i][j] += b.data[i][j];
            a.data[i][j] %= MOD;
        }
    }
    return a;
}

Matrix operator * (int p, Matrix mat) {
    for(int i = 1; i <= mat.n; i++) {
        for(int j = 1; j <= mat.m; i++) {
            mat.data[i][j] *= p;
            mat.data[i][j] %= MOD;
        }
    }
}
 
Matrix pow(Matrix mat, LL p) {
    if(p == 0) {
        Matrix ret(mat.n, mat.m);
        for(int i = 1; i <= mat.n; i++) ret.data[i][i] = 1;
        return ret;
    }
    if(p == 1) return mat;
    Matrix ret = pow(mat * mat, p / 2);
    if(p & 1) ret = ret * mat;
    return ret;
}

int n, m;

int main() {
    int T; scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        Matrix mat(n * m, n * m);
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                int nowx = i - 1, nowy = j - 1, nx, ny;
                char tmp; scanf(" %c", &tmp);
                for(int k = 0; k < 4; k++) {
                    if(k) scanf(" %c", &tmp);
                    scanf("(%d,%d)", &nx, &ny);
                    nx--; ny--;
                    if(i == n && j == m) continue;
                    mat.data[nowx * m + nowy + 1][nx * m + ny + 1] = 1;
                }
                scanf(" %c", &tmp);
            }
        }
        int Q; scanf("%d", &Q);
        while(Q--) {
            int K; scanf("%d", &K);
            Matrix ret = pow(mat, K);
            if(ret.data[1][n * m] == 0) puts("False");
            else {
                int cnt = 0;
                for(int i = 1; i < n * m; i++) {
                    if(ret.data[1][i] != 0) cnt++;
                }
                if(cnt == 0) puts("True");
                else puts("Maybe");
            }
        }
        puts("");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/rolight/p/4049139.html