这两题很类似，都是在补图上搜索
但是由于补图太大我们不能建出来
考虑先从一个点搜，每次搜可以搜的点，
然后维护一个链表，记录当前还没有搜过的点，搜过之后从链表中删除即可
1 type node=record 2 po,next:longint; 3 end; 4 5 var e:array[0..4000010] of node; 6 l:array[0..100010] of node; 7 p,q,ans:array[0..100010] of longint; 8 can,v:array[0..100010] of boolean; 9 s,i,n,m,len,x,y:longint; 10 11 procedure add(x,y:longint); 12 begin 13 inc(len); 14 e[len].po:=y; 15 e[len].next:=p[x]; 16 p[x]:=len; 17 end; 18 19 procedure swap(var a,b:longint); 20 var c:longint; 21 begin 22 c:=a; 23 a:=b; 24 b:=c; 25 end; 26 27 procedure sort(l,r:longint); 28 var i,j,x:longint; 29 begin 30 i:=l; 31 j:=r; 32 x:=ans[(l+r) shr 1]; 33 repeat 34 while ans[i]<x do inc(i); 35 while x<ans[j] do dec(j); 36 if not(i>j) then 37 begin 38 swap(ans[i],ans[j]); 39 inc(i); 40 dec(j); 41 end; 42 until i>j; 43 if l<j then sort(l,j); 44 if i<r then sort(i,r); 45 end; 46 47 procedure del(i:longint); 48 begin 49 l[l[i].po].next:=l[i].next; 50 if l[i].next<>-1 then l[l[i].next].po:=l[i].po; 51 end; 52 53 procedure bfs; 54 var f,r,i,t:longint; 55 begin 56 while l[0].next<>-1 do 57 begin 58 f:=1; 59 r:=1; 60 v[l[0].next]:=true; 61 q[1]:=l[0].next; 62 del(l[0].next); 63 t:=1; 64 while f<=r do 65 begin 66 x:=q[f]; 67 i:=p[x]; 68 while i<>0 do 69 begin 70 can[e[i].po]:=true; 71 i:=e[i].next; 72 end; 73 i:=l[0].next; 74 while i>-1 do 75 begin 76 if not v[i] and not can[i] then 77 begin 78 inc(r); 79 q[r]:=i; 80 del(i); 81 inc(t); 82 v[i]:=true; 83 end; 84 i:=l[i].next; 85 end; 86 i:=p[x]; 87 while i<>0 do 88 begin 89 can[e[i].po]:=false; 90 i:=e[i].next; 91 end; 92 inc(f); 93 end; 94 inc(s); 95 ans[s]:=t; 96 end; 97 end; 98 99 begin 100 readln(n,m); 101 for i:=1 to m do 102 begin 103 readln(x,y); 104 add(x,y); 105 add(y,x); 106 end; 107 for i:=1 to n do 108 begin 109 l[i].po:=i-1; 110 l[i-1].next:=i; 111 end; 112 l[n].next:=-1; 113 bfs; 114 sort(1,s); 115 writeln(s); 116 for i:=1 to s do 117 write(ans[i],' '); 118 writeln; 119 end.
1 type node=record 2 po,next:longint; 3 end; 4 5 var e:array[0..2000010] of node; 6 l:array[0..100010] of node; 7 p:array[0..100010] of longint; 8 can:array[0..100010] of boolean; 9 i,n,m,x,y,len:longint; 10 11 procedure add(x,y:longint); 12 begin 13 inc(len); 14 e[len].po:=y; 15 e[len].next:=p[x]; 16 p[x]:=len; 17 end; 18 19 procedure del(i:longint); 20 begin 21 l[l[i].po].next:=l[i].next; 22 if l[i].next<>-1 then l[l[i].next].po:=l[i].po; 23 end; 24 25 procedure dfs(x:longint); 26 var i,j:longint; 27 begin 28 writeln(x); 29 del(x); 30 i:=p[x]; 31 while i<>0 do 32 begin 33 can[e[i].po]:=true; 34 i:=e[i].next; 35 end; 36 i:=l[0].next; 37 while i>-1 do 38 begin 39 if not can[i] then 40 begin 41 j:=p[x]; 42 while j<>0 do 43 begin 44 can[e[j].po]:=false; 45 j:=e[j].next; 46 end; 47 dfs(i); 48 exit; 49 end; 50 i:=l[i].next; 51 end; 52 end; 53 54 begin 55 readln(n,m); 56 for i:=1 to m do 57 begin 58 readln(x,y); 59 add(x,y); 60 add(y,x); 61 end; 62 for i:=1 to n do 63 begin 64 l[i-1].next:=i; 65 l[i].po:=i-1; 66 end; 67 l[n].next:=-1; 68 dfs(1); 69 end.
