Poj3687 Topology Sorting is not clear yet. Let's look back.

[Question]: N weights are 1 ~ Given the weight comparison relationship between some balls (for example, 2 1 indicates that the second ball is lighter than the first ball), calculate the possible weight of each ball, ans [I] indicates the weight of the I-th ball, which requires that the output be the case with the smallest ans Lexicographic Order.

[Idea]: for the given side of a B, select a value with the smallest sequence number (from N to the largest to the smallest value) for each vertex with an outbound degree of 0 ).

1 # include <iostream> 2 # include <cstdio> 3 # include <cstring> 4 # include <string> 5 # include <algorithm> 6 # include <map> 7 # include <queue> 8 # include <stack> 9 # include <set> 10 # include <cmath> 11 # include <vector> 12 # define INF 0x3f3f3f13 # define INF 0x3fffffffffffffffll14 # define EPS 1e-915 # define PI ACOs (-1.0) 16 using namespace STD; 17 18 int in [202], ANS [202]; 19 bool G [202] [202], vis [202]; 20 priority_qu EUE <int> q; 21 22 int main () 23 {24 int I, T, J, n, m, a, B; 25 scanf ("% d ", & T); 26 while (t --) 27 {28 // getchar (); 29 scanf ("% d", & N, & M ); 30 memset (in, 0, sizeof (in); 31 memset (G, false, sizeof (g); 32 memset (VIS, false, sizeof (VIS )); 33 while (M --) 34 {35 scanf ("% d", & A, & B); 36 IF (G [B] [a]) 37 continue; 38g [B] [a] = true; 39 in [a] ++; 40} 41 for (I = 1; I <= N; I ++) 42 if (in [I] = 0) 43 {44 Q. push (I); 45 vis [I] = true; 46} 47 int num = N; 48 (I = 0; I <n; I ++) // each time a number with an outbound degree of 0 is the smallest 49 {50 int temp; 51 if (Q. empty () 52 break; 53 temp = Q. top (); q. pop (); // printf ("I % d \ n", temp); 54 ans [temp] = num --; 55 for (Int J = 1; j <= N; j ++) 56 If (J! = Temp & G [temp] [J]) 57 {58 in [J] --; 59 If (in [J] = 0 &&! Vis [J]) 60 {61 Q. push (j); 62 vis [J] = true; 63} 64} 66} 67 // printf ("Ans"); 68 if (I <n) 69 printf ("-1 \ n"); 70 else71 {72 printf ("% d", ANS [1]); 73 for (I = 2; I <= N; I ++) 74 printf ("% d", ANS [I]); 75 printf ("\ n"); 76} 77} 78 79}

 

Consider this set of data:

25 41 44 25 33 25 31 44 23 5 output should be: 1 5 3 4 21 3 4 2 5