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HDU 2157 How many ways?? 矩阵
可达矩阵的K次幂便是从i到j走K步能到达的方案数。
注意处理k=0的情况。
#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_backtypedef 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 double DINF = 1e60;const int maxn = 30;const int mod = 1000;struct Matrix { int n, m, data[maxn][maxn]; Matrix(int n = 0, int m = 0): n(n), m(m) { memset(data, 0, sizeof(data)); }};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 pow(Matrix mat, int 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() { while(scanf("%d%d", &n, &m), n) { Matrix mat(n, n); for(int i = 1; i <= m; i++) { int a, b; scanf("%d%d", &a, &b); mat.data[a + 1][b + 1] = 1; } int Q; scanf("%d", &Q); while(Q--) { int a, b, k; scanf("%d%d%d", &a, &b, &k); Matrix ret = pow(mat, k); printf("%d\n", ret.data[a + 1][b + 1]); } } return 0;}
HDU 2157 How many ways?? 矩阵
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