BZOJ1297: [SCOI2009]迷路 矩阵乘法

如果边权都是1，那么直接对当前的邻接矩阵进行T次自乘，答案就是D[1][n]了。

证明：当进行1次自乘时，(D^{1}_{i,j})显然代表从i到j的长度为1的路径条数。假设(D^{k}_{i,j})表示从i到j长度为k的路径条数，(D^{k+1}_{i,j})表示从i到j长度为k+1的路径条数，那么(D^{k + 1} = D^{k} imes D)的(D^{k + 1}_{i, j} = sum_{k = 1}^{n}(D^{k}_{i,k} imes D^{1}_{k,j}))，易知假设成立。得证。

那么对于这道题有什么帮助呢？发现路径长度最长为9，我们可以把每个点拆成至多9个点连成一个链，对于边权为t的边{i, j}连上{I, J - t + 1}，这样每条边的权值都是1，可以做了。

//{HEADS
#define FILE_IN_OUT
#define debug
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <complex>
#include <string>
#define REP(i, j) for (int i = 1; i <= j; ++i)
#define REPI(i, j, k) for (int i = j; i <= k; ++i)
#define REPD(i, j) for (int i = j; 0 < i; --i)
#define STLR(i, con) for (int i = 0, sz = con.size(); i < sz; ++i)
#define STLRD(i, con) for (int i = con.size() - 1; 0 <= i; --i)
#define CLR(s) memset(s, 0, sizeof s)
#define SET(s, v) memset(s, v, sizeof s)
#define mp make_pair
#define pb push_back
#define PL(k, n) for (int i = 1; i <= n; ++i) { cout << k[i] << ' '; } cout << endl
#define PS(k) STLR(i, k) { cout << k[i] << ' '; } cout << endl
using namespace std;
#ifdef debug
#ifndef ONLINE_JUDGE
    const int OUT_PUT_DEBUG_INFO = 1;
#endif
#endif
#ifdef ONLINE_JUDGE
    const int OUT_PUT_DEBUG_INFO = 0;
#endif
#define DG if(OUT_PUT_DEBUG_INFO)
void FILE_INIT(string FILE_NAME) {
#ifdef FILE_IN_OUT
#ifndef ONLINE_JUDGE 
    freopen((FILE_NAME + ".in").c_str(), "r", stdin);
    freopen((FILE_NAME + ".out").c_str(), "w", stdout);

#endif
#endif
}
typedef long long LL;
typedef double DB;
typedef pair<int, int> i_pair;
const int INF = 0x3f3f3f3f;
//}
/*{ 获取字符*/
char gchar() {
    char ret = getchar();
    for(; ret == '
' || ret == '
' || ret == ' '; ret = getchar());
    return ret;
}
/*}*/
const int maxn = 10 * 10;
const int mod = 2009;
int n, T, S;

struct Matrix {
    int d[maxn][maxn];
    Matrix(int t = 0) {
        if(t == 1) {
            REP(i, S) {
                d[i][i] = 1;
            }
        } else {
            CLR(d);
        }
    }
}G, ans;

Matrix operator * (const Matrix & a, const Matrix &b) {
    Matrix ret;
    REP(i, S) {
        REP(j, S) {
            REP(k, S) {
                ret.d[i][j] = (ret.d[i][j] + a.d[i][k] * b.d[k][j] % mod) % mod;
            }
        }
    }
    return ret;
}

int main() {
    FILE_INIT("BZOJ1297");

    scanf("%d %d", &n, &T);
    S = n * 9;
    REP(i, n) {
        REP(j, n) {
            int t = gchar() - '0';
            if(t != 0) {
                G.d[i * 9][j * 9 - t + 1] = 1;
                for(int k = j * 9 - t + 2; k <= j * 9; ++k) {
                    G.d[k - 1][k] = 1;
                }
            }
        }
    }
    /*
    REP(i, S) {
        REP(j, S) {
            printf("%d ", G.d[i][j]);
        }
        puts("");
    }
    */
    for(ans = 1; T; T >>= 1, G = G * G) {
        if(T & 1) {
            ans = ans * G;
        }
    }
    printf("%d
", ans.d[9][S]);


    return 0;
}