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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