做过4010这题其实就水了
把图反向之后直接拓扑排序做即可，我们可以用链表来优化
每个航班的最小起飞序号就相当于在反向图中不用这个点最迟到哪
1 type node=record 2 po,next:longint; 3 end; 4 5 var e:array[0..10010] of node; 6 ans,p,h,suc,du,d,a:array[0..2010] of longint; 7 i,x,y,n,m:longint; 8 9 procedure add(x,y:longint); 10 begin 11 e[i].po:=y; 12 e[i].next:=p[x]; 13 p[x]:=i; 14 end; 15 16 procedure merge(x,y:longint); 17 begin 18 if h[x]=0 then h[x]:=h[y] 19 else begin 20 x:=h[x]; 21 while suc[x]<>0 do x:=suc[x]; 22 suc[x]:=h[y]; 23 end; 24 end; 25 26 function min(wh:longint):longint; 27 var i,x,y,t,b:longint; 28 begin 29 for i:=1 to n do 30 begin 31 du[i]:=d[i]; 32 h[i]:=0; 33 end; 34 for i:=1 to n do 35 if (i<>wh) and (d[i]=0) then 36 begin 37 suc[i]:=h[a[i]]; 38 h[a[i]]:=i; 39 end; 40 t:=n; 41 while t>0 do 42 begin 43 x:=h[t]; 44 if x=0 then exit(t); 45 h[t]:=suc[h[t]]; 46 i:=p[x]; 47 while i<>0 do 48 begin 49 y:=e[i].po; 50 dec(du[y]); 51 if (y<>wh) and (du[y]=0) then 52 begin 53 if a[y]<t then b:=a[y] 54 else b:=t; 55 suc[y]:=h[b]; 56 h[b]:=y; 57 end; 58 i:=e[i].next; 59 end; 60 merge(t-1,t); 61 dec(t); 62 end; 63 exit(0); 64 end; 65 66 procedure work1; 67 var i,x,y,t,b:longint; 68 begin 69 for i:=1 to n do 70 begin 71 du[i]:=d[i]; 72 h[i]:=0; 73 end; 74 for i:=1 to n do 75 if d[i]=0 then 76 begin 77 suc[i]:=h[a[i]]; 78 h[a[i]]:=i; 79 end; 80 t:=n; 81 while t>0 do 82 begin 83 x:=h[t]; 84 ans[t]:=x; 85 h[t]:=suc[h[t]]; 86 i:=p[x]; 87 while i<>0 do 88 begin 89 y:=e[i].po; 90 dec(du[y]); 91 if du[y]=0 then 92 begin 93 if a[y]<t then b:=a[y] 94 else b:=t; 95 suc[y]:=h[b]; 96 h[b]:=y; 97 end; 98 i:=e[i].next; 99 end; 100 merge(t-1,t); 101 dec(t); 102 end; 103 for i:=1 to n-1 do 104 write(ans[i],' '); 105 writeln(ans[n]); 106 end; 107 108 procedure work2; 109 var i:longint; 110 begin 111 for i:=1 to n-1 do 112 write(min(i),' '); 113 writeln(min(n)); 114 end; 115 116 begin 117 readln(n,m); 118 for i:=1 to n do 119 begin 120 read(a[i]); 121 if a[i]>n then a[i]:=n; 122 end; 123 for i:=1 to m do 124 begin 125 readln(x,y); 126 add(y,x); 127 inc(d[x]); 128 end; 129 work1; 130 work2; 131 end.