您的位置:首页 > 其它

ZOJ 3497 Mistwald 矩阵

2014-10-24 19:50 281 查看
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;
}


  
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: