bzoj1297

首先学习是学习矩阵乘法在邻接矩阵的应用
ab两点经过k条边的路径数就等于图的邻接矩阵G的k次幂之后G[a,b]的值
但这道题问的是经过长度为k的路径数
考虑到每条边的长度最长只有9，所以我们把一个点拆成9个点，
i1~i9 顺次相连长度为1，i9为出点
对于长度为k的边eij，连边i9-->j(10-k) 长度为1
这样图中的每条边的长度都变成了1，就相当于求k条边的路径数了

 const mo=;
 var ans,a,b,c:array[..,..] of longint;
     d:array[..] of longint;
     i,j,k,x,n,m:longint;
     ch:char;

 procedure mul;
   var i,j,k:longint;
   begin
     for i:= to n do
       for j:= to n do
       begin
         ans[i,j]:=;
         for k:= to n do
           ans[i,j]:=(ans[i,j]+a[i,k]*b[k,j] mod mo) mod mo;
       end;
    end;

 procedure quick(x:longint);
   var i,j,p,q:longint;
   begin
     j:=;
     while x<> do
     begin
       inc(j);
       d[j]:=x mod ;
       x:=x shr ;
     end;
     fillchar(ans,sizeof(ans),);
     for i:= to n do
       ans[i,i]:=;
     for i:=j downto  do
     begin
       a:=ans;
       b:=ans;
       mul;
       if d[i]= then
       begin
         a:=ans;
         b:=c;
         mul;
       end;
     end;
   end;

 begin
   readln(n,m);
   for i:= to n do
   begin
     for j:= to n do
     begin
       read(ch);
       x:=ord(ch)-;
       if x<> then
       begin
         c[i*,j*-x+]:=;
         for k:=j*-x+ to j*- do
           c[k,k+]:=;
       end;
     end;
     readln;
   end;
   n:=n*;
   quick(m);
   writeln(ans[,n]);
 end.