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BZOJ1297: [SCOI2009]迷路 矩阵乘法

2024-07-14 21:14:27   229人阅读

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，那么\(D^{k + 1} = D^{k} \times D\)的\(D^{k + 1}_{i, j} = \sum_{k = 1}^{n}(D^{k}_{i,k} \times D^{1}_{k,j})\)，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。

 

  1 //{HEADS  2 #define FILE_IN_OUT  3 #define debug  4 #include <cstdio>  5 #include <cstring>  6 #include <cstdlib>  7 #include <cmath>  8 #include <ctime>  9 #include <algorithm> 10 #include <iostream> 11 #include <fstream> 12 #include <vector> 13 #include <stack> 14 #include <queue> 15 #include <deque> 16 #include <map> 17 #include <set> 18 #include <bitset> 19 #include <complex> 20 #include <string> 21 #define REP(i, j) for (int i = 1; i <= j; ++i) 22 #define REPI(i, j, k) for (int i = j; i <= k; ++i) 23 #define REPD(i, j) for (int i = j; 0 < i; --i) 24 #define STLR(i, con) for (int i = 0, sz = con.size(); i < sz; ++i) 25 #define STLRD(i, con) for (int i = con.size() - 1; 0 <= i; --i) 26 #define CLR(s) memset(s, 0, sizeof s) 27 #define SET(s, v) memset(s, v, sizeof s) 28 #define mp make_pair 29 #define pb push_back 30 #define PL(k, n) for (int i = 1; i <= n; ++i) { cout << k[i] << ‘ ‘; } cout << endl 31 #define PS(k) STLR(i, k) { cout << k[i] << ‘ ‘; } cout << endl 32 using namespace std; 33 #ifdef debug 34 #ifndef ONLINE_JUDGE 35     const int OUT_PUT_DEBUG_INFO = 1; 36 #endif 37 #endif 38 #ifdef ONLINE_JUDGE 39     const int OUT_PUT_DEBUG_INFO = 0; 40 #endif 41 #define DG if(OUT_PUT_DEBUG_INFO) 42 void FILE_INIT(string FILE_NAME) { 43 #ifdef FILE_IN_OUT 44 #ifndef ONLINE_JUDGE  45     freopen((FILE_NAME + ".in").c_str(), "r", stdin); 46     freopen((FILE_NAME + ".out").c_str(), "w", stdout); 47  48 #endif 49 #endif 50 } 51 typedef long long LL; 52 typedef double DB; 53 typedef pair<int, int> i_pair; 54 const int INF = 0x3f3f3f3f; 55 //} 56 /*{ 获取字符*/ 57 char gchar() { 58     char ret = getchar(); 59     for(; ret == ‘\n‘ || ret == ‘\r‘ || ret == ‘ ‘; ret = getchar()); 60     return ret; 61 } 62 /*}*/ 63 const int maxn = 10 * 10; 64 const int mod = 2009; 65 int n, T, S; 66  67 struct Matrix { 68     int d[maxn][maxn]; 69     Matrix(int t = 0) { 70         if(t == 1) { 71             REP(i, S) { 72                 d[i][i] = 1; 73             } 74         } else { 75             CLR(d); 76         } 77     } 78 }G, ans; 79  80 Matrix operator * (const Matrix & a, const Matrix &b) { 81     Matrix ret; 82     REP(i, S) { 83         REP(j, S) { 84             REP(k, S) { 85                 ret.d[i][j] = (ret.d[i][j] + a.d[i][k] * b.d[k][j] % mod) % mod; 86             } 87         } 88     } 89     return ret; 90 } 91  92 int main() { 93     FILE_INIT("BZOJ1297"); 94  95     scanf("%d %d", &n, &T); 96     S = n * 9; 97     REP(i, n) { 98         REP(j, n) { 99             int t = gchar() - ‘0‘;100             if(t != 0) {101                 G.d[i * 9][j * 9 - t + 1] = 1;102                 for(int k = j * 9 - t + 2; k <= j * 9; ++k) {103                     G.d[k - 1][k] = 1;104                 }105             }106         }107     }108     /*109     REP(i, S) {110         REP(j, S) {111             printf("%d ", G.d[i][j]);112         }113         puts("");114     }115     */116     for(ans = 1; T; T >>= 1, G = G * G) {117         if(T & 1) {118             ans = ans * G;119         }120     }121     printf("%d\n", ans.d[9][S]);122 123 124     return 0;125 }
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