[Solved] zoj 3497 mistwald matrix fast power

After several months, I re-performed the task and finally got the AC!

The question seems to be graph theory, but the fact is matrix multiplication. According to the discrete mathematics book, bij indicates the number of paths from AI to AJ with a length of K in the new matrix B obtained by K power of the adjacent matrix.

The above method calculates the number of any two-point paths, and this topic requires that the end point cannot have an outbound edge. Based on the sauce purple, it is OK to set all the end points to 0, check whether K steps can reach the end point.

# Include <map> # include <set> # include <list> # include <queue> # include <deque> # include <stack> # include <string> # include <time. h> # include <cstdio> # include <math. h> # include <iomanip> # include <cstdlib> # include <limits. h> # include <string. h> # include <iostream> # include <fstream> # include <algorithm> using namespace STD; # define ll long # define min int_min # define Max int_max # define PI ACOs (-1.0) # define fre Freopen ("input.txt", "r", stdin) # define FF freopen ("output.txt", "W", stdout) int n, m, T; typedef struct node {int mat [30] [30];} MAT; MAT unit, Init; void Init () {char C; scanf ("% d ", & N, & M); getchar (); int I, j, X1, Y1, X2, Y2, X3, Y3, X4, Y4; memset (init. mat, 0, sizeof (init. mat); // clears each time! For (I = 1; I <= N; I ++) for (j = 1; j <= m; j ++) {scanf ("(% d, % d), (% d, % d) ", & X1, & Y1, & X2, & Y2, & X3, & Y3, & X4, & Y4); getchar (); If (I-1) * m + J = N * m) continue; // remove the outdegree init of the destination vertex. mat [(I-1) * m + J] [(x1-1) * m + Y1] = 1; init. mat [(I-1) * m + J] [(x2-1) * m + y2] = 1; init. mat [(I-1) * m + J] [(x3-1) * m + Y3] = 1; init. mat [(I-1) * m + J] [(x4-1) * m + Y4] = 1; unit. mat [(I-1) * m + J] [(I-1) * m + J] = 1;} mat MUL (MAT a, mat B) {int I, j, k; mat c; for (I = 1; I <= T; I ++) for (j = 1; j <= T; j ++) {C. mat [I] [J] = 0; For (k = 1; k <= T; k ++) // you cannot repeat the end point if you forget it at the beginning, to K = T c. mat [I] [J] = C. mat [I] [J] | (. mat [I] [k] & B. mat [k] [J]);} return C;} mat CAL (int K) // matrix binary fast power {mat I = init, ANS = unit; while (k) {If (K & 1) ans = MUL (ANS, I); I = MUL (I, I); k >>=1 ;} return ans ;} int main () {int t; scanf ("% d", & T); bool OK = 0; while (t --) {Init (); // input t = n * m; int num, Q; scanf ("% d", & num); // If (OK) puts (""); OK = 1; int I; while (Num --) {scanf ("% d", & Q); MAT r = CAL (Q); If (R. mat [1] [T] = 0) printf ("false \ n"); else {for (I = 1; I <= T; I ++) {If (R. mat [1] [I]) break;} if (I = T) printf ("True \ n"); else printf ("Maybe \ n ");}} puts ("");} return 0 ;}