ZJU1346 Comparing Your Heroes-count of topological sorting★

Description:

Someone wants to sort the figures of the champion. Enter N, And the next n rows have two strings a B, indicating that a is better than B. The number of different Topology Sorting methods is required. If the sorting output cannot be completed, 0 is returned.

 

Analysis:

AvailableDynamic Planning.

For the topological sorting of N nodes, set the current M nodes with 0 degrees as Zi. SoCurrently, you can select any of the nodes whose m degrees are 0 and the number of sorting methods for the remaining nodes is irrelevant to the previous one.

So you can useNode Set indicates the status. State Transfer: removes a node with a degree of 0 from the current set.

So,Total count of State S = Total number of Zi removed from S by Σ.

When only one node is left, the number of sorting methods is 1.

/* Zju1346 comparing your heroes */# include
 
  
# Include
  
   
# Define N 20 # define M 100001 # define CLR (a) memset (A, 0, sizeof (A) int N; int hash [m]; int a [n] [N]; int DP [m]; int error; int elfhash (char * Key) {unsigned long h = 0; while (* key) {H = (H <4) + * Key ++; unsigned long G = H & 0xf0000000l; If (g) H ^ = G> 24; H & = ~ G;} return H % m;} int ID (char s []) {int H = elfhash (s); If (! Hash [H]) hash [H] = ++ N; return hash [H];} int encode (int e []) {int I, C = 0; for (I = 1; I <= N; I ++) {If (E [I]) c | = 1 <(I-1);} return C ;} int count (int e []) {int c = encode (E); If (DP [c]) return DP [c]; If (error) return 0; int I, j, num = 0; int Z [N], pre, m = 0; for (I = 1; I <= N; I ++) {If (E [I]) {num ++; Pre = 0; For (j = 1; j <= N; j ++) if (E [J] & A [J] [I]) Pre ++; If (! PRE) Z [M ++] = I ;}} if (! M) {// Topology Sorting cannot be completed error = 1; return 0;} If (num = 1) {// only one node DP [c] = 1; return DP [c];} int sum = 0; for (I = 0; I <m; I ++) {e [Z [I] = 0; sum + = count (e); E [Z [I] = 1; if (error) return 0;} DP [c] = sum; return DP [c];} int main () {int m; while (scanf ("% d", & M )! = EOF) {// init n = 0; CLR (hash); CLR (a); CLR (DP); // input int I, J, K; char s [N], t [N]; for (k = 0; k <m; k ++) {scanf ("% S % s", S, T ); I = ID (s); j = ID (t); A [I] [J] = 1;} // DP int e [N]; for (I = 1; I <= N; I ++) E [I] = 1; error = 0; count (E); // output k = encode (E); If (error) printf ("0/N"); else printf ("% d/N", DP [k]);} return 0 ;}