bzoj1875

我们知道，邻接矩阵自乘t次就是经过t条边的路径数
但这道题显然不能这样，因为不能走回头路
于是我们可以构造边的邻接矩阵矩乘即可

 1 const mo=45989;
 2 type way=record
 3        po,fr:longint;
 4      end;
 5      node=array[0..122,0..122] of longint;
 6 
 7 var c,a,w:node;
 8     ed:array[0..200] of way;
 9     x,y,n,m,s,t,e,i,j,ans,len:longint;
10 
11 procedure add(x,y:longint);
12   begin
13     inc(len);
14     ed[len].po:=y;
15     ed[len].fr:=x;
16   end;
17 
18 procedure mul(var c:node;a,b:node);
19   var i,j,k:longint;
20   begin
21     for i:=1 to len do
22       for j:=1 to len do
23       begin
24         c[i,j]:=0;
25         for k:=1 to len do
26           c[i,j]:=(c[i,j]+a[i,k]*b[k,j]) mod mo;
27       end;
28   end;
29 
30 procedure quick(n:longint);
31   begin
32     while n>0 do
33     begin
34       if n mod 2=1 then mul(a,a,w);
35       mul(w,w,w);
36       n:=n shr 1;
37     end;
38   end;
39 
40 begin
41   len:=1;
42   readln(n,m,t,s,e);
43   inc(s);
44   inc(e);
45   for i:=1 to m do
46   begin
47     readln(x,y);
48     inc(x); inc(y);
49     add(x,y);
50     add(y,x);
51   end;
52   for i:=2 to len do
53   begin
54     for j:=2 to len do
55       if (i xor 1<>j) then
56         if ed[i].po=ed[j].fr then w[i,j]:=1;
57     a[i,i]:=1;
58     if ed[i].fr=s then c[1,i]:=1;
59   end;
60   quick(t-1);
61   mul(c,c,a);
62   for i:=2 to len do
63     if ed[i].po=e then ans:=(ans+c[1,i]) mod mo;
64   writeln(ans mod mo);
65 end.

View Code

posted on 2015-04-04 22:48 acphile 阅读(123) 评论(0) 编辑收藏举报