题意:(中问题,题意很简单
思路:a走k步到b,其实就是A^k,ans.mat[a][b]就是答案。
其实就是离散的邻接矩阵那个P(不想证明,逃
#include<cstdio> #include<cmath> #include<iostream> #include<algorithm> #include<vector> #include<stack> #include<cstring> #include<queue> #include<set> #include<string> #include<map> #include <time.h> #define PI acos(-1) using namespace std; typedef long long ll; typedef double db; const int maxn = 20; const int N = 10; const ll maxm = 1e7; const int INF = 0x3f3f3f; const int mod=1000; const ll inf = 1e15 + 5; const db eps = 1e-9; int n, m; struct Matrix{ int mat[maxn][maxn]; Matrix operator*(const Matrix& m)const{ Matrix tmp; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { tmp.mat[i][j]=0; for (int k = 0; k < n; k++) { tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%mod; tmp.mat[i][j]+=mod; tmp.mat[i][j] %= mod; } } } return tmp; } }; Matrix Pow(Matrix m, int k) { Matrix ans; memset(ans.mat , 0 , sizeof(ans.mat)); for (int i=0; i<n; i++) { ans.mat[i][i]=1; } while(k){ if(k&1) ans = ans*m; k >>= 1; m = m*m; } return ans; } void solve() { Matrix tmp; while(scanf("%d%d", &n, &m)!=EOF) { if (!n&&!m) break; memset(tmp.mat, 0, sizeof(tmp.mat)); for (int i=0; i<m; i++) { int u, v; scanf("%d%d", &u, &v); tmp.mat[u][v]=1; } int t; scanf("%d", &t); while(t--) { Matrix hh; int a, b, k; scanf("%d%d%d", &a, &b, &k); hh=Pow(tmp, k); printf("%d\n", hh.mat[a][b]); } } } int main() { int t = 1; //freopen("in.txt", "r", stdin); // scanf("%d", &t); while(t--) solve(); return 0; }