PKU 2594 indicates that a directed acyclic graph is given, and the entire graph is overwritten with the least path. The path allows vertex intersection.
At first glance, this question seems to be the minimum path coverage, but it takes a while to allow vertex intersection. To put it bluntly, since the vertex intersection is allowed, the connectivity should also be passed, so that the connectivity can be changed to the standard minimum path overwrite question. Code Only once, my code is ugly. Code
1 /// Loves DL ~~
2
3 # Include < Iostream >
4 # Include < Algorithm >
5
6 Using Namespace STD;
7
8 Const Int Maxn = 510 ;
9 Bool Gg [maxn] [maxn];
10 Int Link [maxn];
11 Bool S [maxn];
12 Int M, N;
13
14 Bool Find ( Int X)
15 {
16 Int I;
17 For (I = 1 ; I <= N; I ++ ){
18 If (( ! S [I]) && Gg [x] [I]) {
19 S [I] = True ;
20 If (Link [I] = 0 | Find (link [I]) {
21 Link [I] = X;
22 Return True ;
23 }
24 }
25 }
26 Return False ;
27 }
28
29 Int Main ()
30 {
31 Freopen ( " In.txt " , " R " , Stdin );
32 While (Scanf ( " % D " , & N, & M) ! = EOF &&! (M = 0 && N = 0 ))
33 {
34 Memset (Gg, 0 , Sizeof (Gg ));
35 Memset (link, 0 , Sizeof (Link ));
36 Int I, J, K;
37 Int St, E;
38 For (I = 0 ; I < M; I ++ )
39 {
40 Scanf ( " % D " , & St, & E );
41 Gg [st] [E] = True ;
42 }
43 /// /Floyd evaluate the passing Closure
44 For (K = 1 ; K <= N; k ++ )
45 {
46 For (I = 1 ; I <= N; I ++ )
47 {
48 For (J = 1 ; J <= N; j ++ )
49 {
50 If (Gg [I] [k] && Gg [k] [J] &&! Gg [I] [J])
51 Gg [I] [J] = True ;
52 }
53 }
54 }
55 /// /Find the minimum path Overwrite
56 Int Ans = 0 ;
57 For (I = 1 ; I <= N; I ++ )
58 {
59 Memset (S, False , Sizeof (S ));
60 If (Find (I ))
61 {
62 Ans ++ ;
63 }
64 }
65 Printf ( " % D \ n " , N - Ans );
66 }
67
68 Return 0 ;
69
