这是yhc大牛矩乘论文上的题目，那里面分析得很清楚了
这里就说说我一开始错的地方，我一开始处理每个时刻能通过的邻接矩阵的时候
把以不能访问的点i为起点和终点的都标记为0了，实际上只能标记以i为终点的边即可
很好理解，是自己太脑残了
1 const mo=10000; 2 var b,c,g,f:array[0..51,0..51] of longint; 3 v:array[0..51,0..51] of boolean; 4 w,d:array[0..51] of longint; 5 a:array[0..15,0..51,0..51] of longint; 6 fish,i,j,k,x,y,n,m,s,e,t:longint; 7 8 procedure work(i:longint); 9 var j,k:longint; 10 begin 11 a[i]:=f; 12 for j:=1 to n do 13 if v[j,i] then 14 begin 15 for k:=1 to n do 16 a[i,k,j]:=0; 17 end; 18 end; 19 20 procedure mul(p:longint); 21 var i,j,k:longint; 22 begin 23 b:=c; 24 for i:=1 to n do 25 for j:=1 to n do 26 begin 27 c[i,j]:=0; 28 for k:=1 to n do 29 c[i,j]:=(c[i,j]+b[i,k]*a[p,k,j] mod mo) mod mo; 30 end; 31 end; 32 33 procedure time; 34 var i,j,k:longint; 35 begin 36 for i:=1 to n do 37 for j:=1 to n do 38 begin 39 c[i,j]:=0; 40 for k:=1 to n do 41 c[i,j]:=(c[i,j]+g[i,k]*f[k,j] mod mo) mod mo; 42 end; 43 end; 44 45 procedure quick(x:longint); 46 var i,j:longint; 47 begin 48 j:=0; 49 while x>0 do 50 begin 51 inc(j); 52 d[j]:=x mod 2; 53 x:=x shr 1; 54 end; 55 fillchar(c,sizeof(c),0); 56 for i:=1 to n do 57 c[i,i]:=1; 58 for i:=j downto 1 do 59 begin 60 g:=c; 61 f:=c; 62 time; 63 if d[i]=1 then 64 begin 65 g:=c; 66 f:=b; 67 time; 68 end; 69 end; 70 end; 71 72 begin 73 readln(n,m,s,e,t); 74 inc(s); 75 inc(e); 76 for i:=1 to m do 77 begin 78 readln(x,y); 79 inc(x); 80 inc(y); 81 f[x,y]:=1; 82 f[y,x]:=1; 83 end; 84 readln(fish); 85 for i:=1 to fish do 86 begin 87 read(y); 88 for j:=0 to y-1 do 89 begin 90 read(x); 91 inc(x); 92 k:=j; 93 while k<=12 do 94 begin 95 v[x,k]:=true; 96 k:=k+y; 97 end; 98 end; 99 end; 100 for i:=0 to 12 do 101 work(i); 102 if t<12 then 103 begin 104 c:=a[1]; 105 for i:=2 to t do 106 mul(i); 107 end 108 else begin 109 c:=a[1]; 110 for i:=2 to 12 do 111 mul(i); 112 b:=c; 113 quick(t div 12); 114 for i:=1 to t mod 12 do 115 mul(i); 116 end; 117 writeln(c[s,e]); 118 end.
