CH Round #51 - Shinrein祭 #1

A.Phorni

题目：http://www.contesthunter.org/contest/CH%20Round%20%2351%20-%20Shinrein祭%20%231/Phorni

没做。。。

B.Arietta

题目：http://www.contesthunter.org/contest/CH%20Round%20%2351%20-%20Shinrein祭%20%231/Arietta

想到了网络流，所以每次暴力算出哪些点能被弹奏，就从这次弹奏向哪些点连容量为1的边

最后由s向所有力度连容量为该力度最多能被弹多少次的边，由每个节点连t容量为1的边

求最大流，得到了暴力分30分。正解还不会

代码：

 const inf=maxlongint;maxn=+;maxm=;
 type node=record
      go,next,v:longint;
      end;
 var  tot,i,n,m,maxflow,l,r,s,t,x,y,tot2:longint;
      h,head,head2,q,cur,v:array[..maxn] of longint;
      e:array[..maxm] of node;
      e2:array[..maxn] of node;
      function min(x,y:longint):longint;
       begin
       if x<y then exit(x) else exit(y);
       end;
 procedure ins(x,y,z:longint);
  begin
  inc(tot);
  e[tot].go:=y;e[tot].v:=z;e[tot].next:=head[x];head[x]:=tot;
  end;
 procedure insert(x,y,z:longint);
  begin
  ins(x,y,z);ins(y,x,);
  end;
 function bfs:boolean;
  var i,x,y:longint;
  begin
  fillchar(h,sizeof(h),);
  l:=;r:=;q[]:=s;h[s]:=;
  while l<r do
   begin
   inc(l);
   x:=q[l];
   i:=head[x];
   while i<> do
    begin
    y:=e[i].go;
    if (e[i].v<>) and (h[y]=) then
     begin
      h[y]:=h[x]+;
      inc(r);q[r]:=y;
     end;
    i:=e[i].next;
    end;
   end;
  exit (h[t]<>);
  end;
 function dfs(x,f:longint):longint;
  var i,y,used,tmp:longint;
  begin
  if x=t then exit(f);
  used:=;
  i:=cur[x];
  while i<> do
   begin
   y:=e[i].go;
   if (h[y]=h[x]+) and (e[i].v<>) then
    begin
    tmp:=dfs(y,min(e[i].v,f-used));
    dec(e[i].v,tmp);if e[i].v<> then cur[x]:=i;
    inc(e[i xor ].v,tmp);
    inc(used,tmp);
    if used=f then exit(f);
    end;
   i:=e[i].next;
   end;
  if used= then h[x]:=-;
  exit(used);
  end;
 procedure dinic;
  var i:longint;
  begin
  while bfs do
   begin
   for i:=s to t do cur[i]:=head[i];
   inc(maxflow,dfs(s,inf));
   end;
  end;
 procedure insert2(x,y:longint);
  begin
  inc(tot2);
  e2[tot2].go:=y;e2[tot2].next:=head2[x];head2[x]:=tot2;
  end;
 procedure init;
  begin
    tot:=;
    readln(n,m);
    for i:= to n do begin read(x);insert2(x,i);end;readln;
    for i:= to n do read(v[i]);readln;
  end;
 procedure dfss(x:longint);
  var j,y:longint;
  begin
  if (v[x]>=l) and (v[x]<=r) then insert(i,x,);
  j:=head2[x];
  while j<> do
   begin
     y:=e2[j].go;
     dfss(y);
     j:=e2[j].next;
   end;
  end;

 procedure main;
  begin
    s:=;t:=n+m+;
    for i:= to n do insert(i,t,);
    for i:=n+ to n+m do
     begin
       readln(l,r,x,y);
       insert(s,i,y);
       dfss(x);
     end;
    maxflow:=;
    dinic;
    writeln(maxflow);
  end;

 begin
   init;
   main;
 end.                       

C。Falsita

题目：http://www.contesthunter.org/contest/CH%20Round%20%2351%20-%20Shinrein祭%20%231/Falsita

我下午刚做了POI的大都市，然后看到这题的对子树的修改就想到了用dfs序+线段树懒惰标记的做法，

最后回答的时候我是O(该点分叉数）求出总数，再用同样时间复杂度的时间求出总代价，使用子树和算的

我想出题人的要求回答询问的该点的分叉树一定很多，就会卡我到超时，结果真是的。。。

于是我又很愉快的拿到了30分。。。正解还不会

代码：

 const maxn=+;
 type node=record
      go,next:longint;
      end;
      node2=record
      l,r,lch,rch,mid,tag,sum:int64;
      end;
 var  e:array[..*maxn] of node;
      t:array[..*maxn] of node2;
      head,l,r,s,a,b:array[..*maxn] of longint;
      i,n,m,x,y,tot,clock:longint;
      ch:char;
      procedure insert(x,y:longint);
       begin
         inc(tot);
         e[tot].go:=y;e[tot].next:=head[x];head[x]:=tot;
       end;
 procedure dfs(x:longint);
  var i,y:longint;
  begin
    inc(clock);l[x]:=clock;a[clock]:=x;
    i:=head[x];
    while i<> do
     begin
      y:=e[i].go;
      dfs(y);
      i:=e[i].next;
     end;
    inc(clock);r[x]:=clock;a[clock]:=x;
  end;
 procedure pushup(k:longint);
  begin
    with t[k] do
     begin
      sum:=t[lch].sum+t[rch].sum;
     end;
  end;
 procedure build(k,x,y:longint);
  begin
    with t[k] do
     begin
      l:=x;r:=y;mid:=(l+r)>>;lch:=k<<;rch:=k<<+;tag:=;
      if l=r then begin sum:=b[a[l]];exit;end;
      build(lch,l,mid);build(rch,mid+,r);
      pushup(k);
     end;
  end;
 procedure update(k:longint;val:int64);
  begin
    with t[k] do
     begin
      inc(tag,val);
      inc(sum,(r-l+)*val);
     end;
  end;
 procedure pushdown(k:longint);
  begin
    with t[k] do
     begin
      if tag= then exit;
      update(lch,tag);update(rch,tag);
      tag:=;
     end;
  end;
 procedure change(k,x,y:longint);
  begin
    with t[k] do
     begin
      if l=r then begin inc(sum,y);exit;end;
      pushdown(k);
      if x<=mid then change(lch,x,y) else change(rch,x,y);
      pushup(k);
     end;
  end;
 procedure change2(k,x,y:longint;z:int64);
  begin
    with t[k] do
     begin
      if (l=x) and (r=y) then
       begin
         update(k,z);exit;
       end;
      pushdown(k);
      if y<=mid then change2(lch,x,y,z)
      else if x>mid then change2(rch,x,y,z)
      else
         begin
           change2(lch,x,mid,z);
           change2(rch,mid+,y,z);
         end;
      pushup(k);
     end;
  end;
 function query(k,x,y:longint):int64;
  begin
    with t[k] do
     begin
      if (l=x) and (r=y) then exit(sum);
      pushdown(k);
      if y<=mid then exit(query(lch,x,y))
      else if x>mid then exit(query(rch,x,y))
      else exit(query(lch,x,mid)+query(rch,mid+,y));
     end;
  end;
 procedure init;
  begin
    readln(n,m);
    for i:= to n do begin read(x);insert(x,i);end;readln;
    for i:= to n do read(b[i]);readln;
    dfs();
    build(,,*n);
    for i:= to n do s[i]:=(r[i]-l[i]+)>>;
  end;
 procedure getans;
  var ans:double;
      a,b,c:array[..maxn] of int64;
      i,x,y,cnt:longint;
      tot:int64;
  begin
    readln(x);a[]:=x;b[]:=s[x];c[]:=trunc(query(,l[x],r[x])/);
    cnt:=;
    i:=head[x];
    while i<> do
     begin
      y:=e[i].go;
      inc(cnt);
      a[cnt]:=y;b[cnt]:=s[y];c[cnt]:=trunc(query(,l[y],r[y])/);
      dec(c[],c[cnt]);
      i:=e[i].next;
     end;
    tot:=b[]-;
    for i:= to cnt do inc(tot,b[i]*(b[]-b[i]));
    tot:=tot>>;
    ans:=c[]*(b[]-)/tot;
    for i:= to cnt do ans:=ans+c[i]*(b[]-b[i])/tot;
    writeln(ans::);
  end;
 procedure main;
  begin
    for i:= to m do
     begin
      read(ch);
      case ch of
      'S':begin
           readln(x,y);
           change(,l[x],y);change(,r[x],y);
          end;
      'M':begin
           readln(x,y);
           change2(,l[x],r[x],y);
          end;
      'Q':getans;
      end;
     end;
  end;
 begin
   init;
   main;
 end.