HDOJ3341Lost's revenge（AC自动机，DP）

Posted 2020-10-09 myx12345

题意：给出一个n个模式串，一个目标串，问把目标串重新排位最多能产生多少个模式串，可以重叠且所有串只包含A C G T。 

n<=10,len[i]<=10

len(s)<=40

Cas<=30

思路：TLE，估计被卡常了

可以将题意理解为重新构造一个ACGT个数都与原目标串相同的新串，则目标串中有用的信息只有ACGT个数

建出Trie图，跑一遍AC自动机，再用一个二维dp[i,j]表示trie上i号节点，字母使用情况为j的模式串最多出现次数

其中j为状态压缩量，是一个四维的量压成一维，因为不压的话会MLE

dp[v,h(a,b,c,d)]=max(dp[v,h(a,b,c,d)],dp[u,h(a-1,b,c,d)]+size[v])等四种转移

其中v=map[u,i]，size[v]表示trie上到v节点路径出现次数之和

边界条件dp[1,h(0,0,0,0)]=1

  1 var map:array[1..600,1..4]of longint;
  2     q,a,size,fa:array[0..16000]of longint;
  3     hash:array[0..40,0..40,0..40,0..40]of longint;
  4     dp:array[0..600,0..16000]of longint;
  5     ch:string;
  6     n,cnt,i,j,k,x,t,y,ans,len,s,cas,a1,c1,g1,t1,u,s1,v,s2,num:longint;
  7     flag:boolean;
  8 
  9 function min(x,y:longint):longint;
 10 begin
 11  if x<y then exit(x);
 12  exit(y);
 13 end;
 14 
 15 function max(x,y:longint):longint;
 16 begin
 17  if x>y then exit(x);
 18  exit(y);
 19 end;
 20 
 21 procedure build;
 22 var i,u:longint;
 23 begin
 24  u:=1;
 25  for i:=1 to len do
 26  begin
 27   if map[u,a[i]]=0 then
 28   begin
 29    inc(cnt); map[u,a[i]]:=cnt;
 30   end;
 31   u:=map[u,a[i]];
 32  end;
 33  inc(size[u]);
 34 end;
 35 
 36 procedure acauto;
 37 var t,w,i,p,son,u:longint;
 38 begin
 39  t:=0; w:=1; q[1]:=1;
 40  while t<w do
 41  begin
 42   inc(t); u:=q[t];
 43   size[u]:=size[u]+size[fa[u]];
 44   for i:=1 to 4 do
 45    if map[u,i]>0 then
 46    begin
 47     son:=map[u,i];
 48     p:=fa[u];
 49     if u=1 then fa[son]:=1
 50      else fa[son]:=map[p,i];
 51     inc(w); q[w]:=son;
 52    end
 53     else
 54     begin
 55      p:=fa[u];
 56      if u=1 then map[u,i]:=1
 57       else map[u,i]:=map[p,i];
 58     end;
 59  end;
 60 end;
 61 
 62 begin
 63  assign(input,‘hdoj3341.in‘); reset(input);
 64  assign(output,‘hdoj3341.out‘); rewrite(output);
 65  while not eof do
 66  begin
 67   readln(n);
 68   if n=0 then break;
 69   inc(cas);
 70   for i:=1 to cnt do
 71   begin
 72    size[i]:=0; fa[i]:=0;
 73   end;
 74   for i:=1 to cnt do
 75    for j:=1 to 4 do map[i,j]:=0;
 76   for i:=0 to a1 do
 77    for j:=0 to c1 do
 78     for k:=0 to g1 do
 79      for x:=0 to t1 do hash[i,j,k,x]:=0;
 80 
 81   cnt:=1;
 82   for i:=1 to n do
 83   begin
 84    readln(ch);
 85    len:=length(ch);
 86    for j:=1 to len do
 87    begin
 88     case ch[j] of
 89      ‘A‘:a[j]:=1;
 90      ‘C‘:a[j]:=2;
 91      ‘G‘:a[j]:=3;
 92      ‘T‘:a[j]:=4;
 93     end;
 94    end;
 95    build;
 96   end;
 97   acauto;
 98   readln(ch);
 99   len:=length(ch); a1:=0; c1:=0; g1:=0; t1:=0;
100   for i:=1 to len do
101   begin
102    case ch[i] of
103     ‘A‘:inc(a1);
104     ‘C‘:inc(c1);
105     ‘G‘:inc(g1);
106     ‘T‘:inc(t1);
107    end;
108   end;
109   num:=0;
110   for i:=0 to a1 do
111    for j:=0 to c1 do
112     for k:=0 to g1 do
113      for x:=0 to t1 do
114      begin
115       inc(num); hash[i,j,k,x]:=num;
116      end;
117 
118  for i:=1 to cnt do
119   for j:=0 to a1 do
120    for k:=0 to c1 do
121     for x:=0 to g1 do
122      for y:=0 to t1 do dp[i,hash[j,k,x,y]]:=-1;
123   dp[1,hash[0,0,0,0]]:=0;
124 
125   ans:=0;
126   for i:=0 to a1 do
127    for j:=0 to c1 do
128     for k:=0 to g1 do
129      for x:=0 to c1 do
130      begin
131       if i+j+k+x=0 then continue;
132       s1:=hash[i,j,k,x];
133       for u:=1 to cnt do
134        for t:=1 to 4 do
135        begin
136         v:=map[u,t];
137         s2:=0;
138         if (t=1)and(i>=1) then s2:=hash[i-1,j,k,x]
139         else if (t=2)and(j>=1) then s2:=hash[i,j-1,k,x]
140          else if (t=3)and(k>=1) then s2:=hash[i,j,k-1,x]
141           else  if (t=4)and(x>=1) then s2:=hash[i,j,k,x-1];
142         if s2=0 then continue;
143         if dp[u,s2]=-1 then continue;
144         dp[v,s1]:=max(dp[v,s1],dp[u,s2]+size[v]);
145         ans:=max(ans,dp[v,s1]);
146        end;
147      end;
148   writeln(‘Case ‘,cas,‘: ‘,ans);
149  end;
150  close(input);
151  close(output);
152 end.