【NOIP2017】宝藏（状压DP）

题意：

思路：n<=12，考虑状压DP

生成树中深度相同的点可以一次性转移完毕

设dp[sta,i]为已转移完sta状态的点，当前深度为i的最小花费

dp[sta or v,i+1]=min(dp[sta,i]+f[sta,v]*(i+1))，其中v是sta关于全集(1<<n)-1的补集v1的一个子集，这一步需要枚举子集

考场上写的O(3^n*n^2)，没有预处理f[sta,v]而是每次都算了一遍，有进一步优化的空间

 const max=;
 var dp,dis:array[..,..]of int64;
     f:array[..,..]of int64;
     n,m,i,j,x,y,z,k,maxs,v,v1:longint;
     s,ans:int64;

 function min(x,y:int64):int64;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 begin
  assign(input,'treasure.in'); reset(input);
  assign(output,'treasure.out'); rewrite(output);
  readln(n,m);
  for i:= to n do
   for j:= to n do f[i,j]:=<<;
  for i:= to n do f[i,i]:=;
  for i:= to m do
  begin
   readln(x,y,z);
   f[x,y]:=min(f[x,y],z);
   f[y,x]:=min(f[y,x],z);
  end;
  maxs:=(<<n)-;
  for i:= to maxs do
   for j:= to n do
    if i and (<<(j-))= then
    begin
     dis[i,j]:=<<;
     for k:= to n do
      if (j<>k)and(i and (<<(k-))>) then dis[i,j]:=min(dis[i,j],f[j,k]);
    end;

  m:=maxs;
  for i:= to maxs do
   for j:= to n+ do dp[i,j]:=<<;
  for i:= to n do dp[<<(i-),]:=;
  for i:= to n do
   for j:= to maxs do
   begin
    v:=j xor m; v1:=v;
    while v> do
    begin
     s:=;
     for k:= to n do
      if v and (<<(k-))> then
      begin
       s:=s+dis[j,k];
       if s>=(<<) then break;
      end;
     if s<(<<) then
      dp[j or v,i+]:=min(dp[j or v,i+],dp[j,i]+s*(i+));
     v:=v1 and (v-);
    end;
   end;
  ans:=<<;
  for i:= to n+ do ans:=min(ans,dp[maxs,i]);
  if n= then ans:=;
  writeln(ans);

  close(input);
  close(output);
 end.

O(3^n*n)，预处理两个值

d[sta,i]      已取sta状态中的点到i点的最小值 预处理O（2^n*n^2）

f[x,y]      x状态中的点和y状态中的所有点连接最小长度之和=f[x,y-lowbit(y)]+d[x,z]，z表示y中最后一个1的位置 预处理O（4^n） 需要保证x与y没有交集

 const max=;
 var dp,d:array[..,..]of int64;
     dis:array[..,..]of int64;
     f:array[..,..]of int64;
     num:array[..]of longint;
     n,m,i,j,x,y,z,k,maxs,v,v1:longint;
     s,ans:int64;

 function min(x,y:int64):int64;
 begin
  if x<y then exit(x);
  exit(y);
 end;

 function lowbit(x:longint):longint;
 begin
  exit(x and (-x));
 end;

 begin
  assign(input,'treasure.in'); reset(input);
  assign(output,'treasure.out'); rewrite(output);
  readln(n,m);
  for i:= to n do
   for j:= to n do f[i,j]:=<<;
  for i:= to n do f[i,i]:=;
  for i:= to m do
  begin
   readln(x,y,z);
   f[x,y]:=min(f[x,y],z);
   f[y,x]:=min(f[y,x],z);
  end;

  maxs:=(<<n)-;
  for i:= to maxs do
   for j:= to n do
    if i and (<<(j-))= then
    begin
     d[i,j]:=<<;
     for k:= to n do
      if (j<>k)and(i and (<<(k-))>) then d[i,j]:=min(d[i,j],f[j,k]);
    end;

  m:=maxs;
  for i:= to  do num[<<(i-)]:=i;
  for i:= to maxs do
   for j:= to maxs do dis[i,j]:=<<;
  for i:= to maxs do dis[,i]:=;
  for i:= to maxs do dis[i,]:=;

  for i:= to maxs do
   for j:= to maxs do
    if i and j= then
    begin
     x:=num[lowbit(j)];
     dis[i,j]:=dis[i,j-lowbit(j)]+d[i,x];
     if dis[i,j]>(<<) then dis[i,j]:=<<;
    end;

  for i:= to maxs do
   for j:= to n+ do dp[i,j]:=<<;
  for i:= to n do dp[<<(i-),]:=;
  for i:= to n do
   for j:= to maxs do
   begin
    v:=j xor m; v1:=v;
    while v> do
    begin
     if dis[j,v]<<< then
      dp[j or v,i+]:=min(dp[j or v,i+],dp[j,i]+dis[j,v]*(i+));
     v:=v1 and (v-);
    end;
   end;

  ans:=<<;
  for i:= to n+ do ans:=min(ans,dp[maxs,i]);
  if n= then ans:=;
  writeln(ans);

  close(input);
  close(output);
 end.