bzoj1444

显然自动机上高斯消元
根据AC自动机上的转移可以列出一系列方程
但这个样的方程解出来全0是一组解
说明有线性组合的情况
考虑除非没人能赢，否则每个人赢的概率和为1
那么我们只要把原来的第一个方程换成这个即可

 var ans,d:array[..] of double;
     a:array[..,..] of double;
     w,q,f:array[..] of longint;
     trie:array[..,..] of longint;
     v:array[..] of boolean;
     j,k,i,n,m,l,x,y,t:longint;
     s:string;

 procedure swap(var a,b:double);
   var c:double;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure ac;
   var h,r,x,y,i:longint;
   begin
     h:=;
     r:=;
     for i:= to m do
       if trie[,i]> then
       begin
         inc(r);
         q[r]:=trie[,i];
       end;

     while h<=r do
     begin
       x:=q[h];
       for i:= to m do
         if trie[x,i]> then
         begin
           y:=trie[x,i];
           inc(r);
           q[r]:=y;
           j:=f[x];
           while (j>) and (trie[j,i]=) do j:=f[j];
           f[y]:=trie[j,i];
         end;
       inc(h);
     end;
   end;

 procedure gauss;
   var i,j,k,p,u:longint;
       c:double;
   begin
     u:=;
     for i:= to t do
     begin
       p:=u;
       for j:=u+ to t do
         if abs(a[j,i])>abs(a[p,i]) then p:=j;

       if (p=u) and (a[p,i]=) then continue;
       if p<>u then
       begin
         for j:= to t+ do
           swap(a[u,j],a[p,j]);
       end;
       for j:=u+ to t do
         if a[j,i]<> then
         begin
           c:=a[j,i]/a[u,i];
           for k:= to t+ do
             a[j,k]:=a[j,k]-a[u,k]*c;
         end;
       inc(u);
     end;
     for i:=t downto  do
     begin
       if a[i,i]= then continue;
       for j:=i+ to t do
         a[i,t+]:=a[i,t+]-ans[j]*a[i,j];
       ans[i]:=a[i,t+]/a[i,i];
     end;
   end;

 begin
   readln(n,l,m);
   for i:= to m do
   begin
     readln(x,y);
     d[i]:=x/y;
   end;
   for i:= to n do
   begin
     readln(s);
     j:=;
     for k:= to l do
     begin
       x:=ord(s[k])-;
       if trie[j,x]= then
       begin
         inc(t);
         trie[j,x]:=t;
       end;
       j:=trie[j,x];
     end;
     w[i]:=j;
     v[j]:=true;
   end;
   ac;
   for i:= to t do
   begin
     a[i,i]:=-;
     if not v[i] then
     begin
       for k:= to m do
       begin
         j:=i;
         while (j>) and (trie[j,k]=) do j:=f[j];
         a[trie[j,k],i]:=a[trie[j,k],i]+d[k];
       end;
     end;
   end;
   for i:= to t do
     if not v[i] then a[,i]:= else a[,i]:=;
   a[,t+]:=;
   gauss;
   for i:= to n do
     if (ans[w[i]]>) and (ans[w[i]]<=) then
       writeln(ans[w[i]]::)
     else writeln('0.00');
 end.