这不是floyd
tarjan缩点，同一个强联通分量中，元素只有一个就不需要连边，否则就连成一个环，ans+=元素个数
处理强联通分量直接连边有点麻烦，先把所有的边都连上，并记录连边的出发点和终点(l[num],r[num]),ans+=num
用p[i][j]来记录强联通分量i到j的走法总数,用floyd计算,p[i][j]+=p[i][k]*p[k][j] (>0),然后再枚举每一条连着的边,看能否删掉,能删掉的标志是p[i][j]>1,于是ans--,p[i][j]--
附代码:
var n,m,o,i,ii,j,k,v,x,y,tot,top,co,num,ans:longint;
head,next,vet:array[0..40005] of longint;
color,st,count,dfn,low,d,l,r:array[0..300] of longint;
map,p:array[0..205,0..205] of longint;
b:array[0..205] of boolean;
function min(a,b:longint):longint;
begin if A>b then exit(b);exit(a);end;
procedure tarjan(x:longint);
var i,v:longint;
begin
inc(num);low[x]:=num;dfn[x]:=num;
inc(top);st[top]:=x;
i:=head[x];
while i<>0 do begin
v:=vet[i];
if dfn[v]=0 then begin
tarjan(v);
low[x]:=min(low[x],low[v]);
end
else if color[v]=0 then low[x]:=min(low[x],dfn[v]);
i:=next[i];
end;
if dfn[x]=low[x] then begin
inc(co);
while st[top]<>x do begin
color[st[top]]:=co;
inc(count[co]);
dec(top);
end;
inc(count[co]);
color[st[top]]:=co;
dec(top);
end;
end;
begin
readln(o);
for o:=o downto 1 do begin
readln(n);
tot:=0;num:=0;co:=0;top:=0;ans:=0;
for i:=1 to n do begin b[i]:=false;dfn[i]:=0;low[i]:=0;count[i]:=0;color[i]:=0;head[i]:=0;d[i]:=0;end;
for i:=1 to n do
for j:=1 to n do begin
read(x);map[i,j]:=x;p[i,j]:=0;
if i=j then continue;
if x=1 then begin
inc(tot);
next[tot]:=head[i];
head[i]:=tot;
vet[tot]:=j;
end;
end;
for i:=1 to n do if dfn[i]=0 then tarjan(i);
for i:=1 to n do begin
ii:=head[i];
while ii<>0 do begin
v:=vet[ii];
if color[v]<>color[i] then
inc(d[color[v]]);
ii:=next[ii];
end;
end;
num:=0;
for i:=1 to n do
for j:=1 to n do
if (map[i][j]=1)and(color[i]<>color[j]) then begin
inc(num);
l[num]:=color[i];
r[num]:=color[j];
inc(p[color[i]][color[j]]);
end;
for k:=1 to co do
for i:=1 to co do
for j:=1 to co do
if (p[i,k]>0)and(p[k][j]>0) then inc(p[i,j],p[i,k]*p[k][j]);
ans:=ans+num;
for i:=1 to num do
if p[l[i]][r[i]]=1 then continue else begin dec(ans);dec(p[l[i]][r[i]]);end;
for i:=1 to co do
if count[i]>1 then ans:=ans+count[i];
writeln(ans);
end;
end.