Determine the position of the matchTime
limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 17134 Accepted Submission (s): 6789
Problem description has N race teams (1<=n<=500), numbered 1, 2. 3, .....
。 n Play the game. At the end of the game, the Referee Committee will rank all the participating teams from the point of travel, but now the referee Committee cannot get the results of each team directly. Just know the results of each game, that is P1 win P2, with P1,p2 said, ranked P1 before P2.
Now ask you to compile the program to determine the rankings.
Input inputs have several groups, the first behavior in each group is two number N (1<=n<=500), M. n indicates the number of teams. m indicates that the input data of the M row is followed. In the next M-row data, there are also two integers per line p1,p2 means that the P1 team won the P2 team.
Output gives a ranking that meets the requirements. There is a space between the queue numbers at the time of the output, and no space after the last.
Other Notes: Qualifying rankings may not be unique, at which point a small number of teams are required to output, and the input data is guaranteed to be correct. That is, the input data ensures that there must be a ranking that meets the requirements.
Sample Input
4 31 22) 34 3
Sample Output
1 2 4 3
Note two points:
(1): There may be a heavy edge in the input. Do not consider the adjacency table processing.
(2): The topic requires the output of small number in front, so with priority queue to store
#include <cstdio> #include <cstring> #include <algorithm> #include <queue>using namespace std; struct node {int u, V, next;}; int N, m;node edge[550 * 550];int in[550];int head[550], cnt;int sum[550];void init () {cnt = 0; Memset (Head,-1, sizeof (head)); memset (in, 0, sizeof (in)); memset (sum, 0, sizeof (sum));} void Add (int u, int v) {edge[cnt] = {u, V, Head[u]}; Head[u] = cnt++;} void Topsort () {int ans = 0; Priority_queue<int,vector<int>,greater<int> >q; for (int i = 1; I <= n; ++i) {if (in[i] = = 0) Q.push (i); } while (!q.empty ()) {int u = q.top (); sum[ans++] = u; Q.pop (); for (int i = head[u]; i =-1; i = edge[i].next) {int v = EDGE[I].V; in[v]--; if (in[v] = = 0) q.push (v); }} for (int i = 0; i < ans; ++i) {if (!i) printf ("%d", sum[i]); else printf ("%d", sum[i]); } printf ("\ n");} int main () {while (scanf ("%d%d", &n, &m)! = EOF) {init (); while (m--) {int A, B; scanf ("%d%d", &a, &b); Add (A, b); in[b]++; } topsort (); } return 0;}
HDU 1285--to determine the rank of the contest "topological sorting && adjacency Table Implementation "