【HDOJ3341】Lost's revenge（AC自动机，DP）

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

var map:array[1..600,1..4]of longint;
    q,a,size,fa:array[0..16000]of longint;
    hash:array[0..40,0..40,0..40,0..40]of longint;
    dp:array[0..600,0..16000]of longint;
    ch:string;
    n,cnt,i,j,k,x,t,y,ans,len,s,cas,a1,c1,g1,t1,u,s1,v,s2,num:longint;
    flag:boolean;

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

function max(x,y:longint):longint;
begin
 if x>y then exit(x);
 exit(y);
end;

procedure build;
var i,u:longint;
begin
 u:=1;
 for i:=1 to len do
 begin
  if map[u,a[i]]=0 then
  begin
   inc(cnt); map[u,a[i]]:=cnt;
  end;
  u:=map[u,a[i]];
 end;
 inc(size[u]);
end;

procedure acauto;
var t,w,i,p,son,u:longint;
begin
 t:=0; w:=1; q[1]:=1;
 while t<w do
 begin
  inc(t); u:=q[t];
  size[u]:=size[u]+size[fa[u]];
  for i:=1 to 4 do
   if map[u,i]>0 then
   begin
    son:=map[u,i];
    p:=fa[u];
    if u=1 then fa[son]:=1
     else fa[son]:=map[p,i];
    inc(w); q[w]:=son;
   end
    else
    begin
     p:=fa[u];
     if u=1 then map[u,i]:=1
      else map[u,i]:=map[p,i];
    end;
 end;
end;

begin
 assign(input,'hdoj3341.in'); reset(input);
 assign(output,'hdoj3341.out'); rewrite(output);
 while not eof do
 begin
  readln(n);
  if n=0 then break;
  inc(cas);
  for i:=1 to cnt do
  begin
   size[i]:=0; fa[i]:=0;
  end;
  for i:=1 to cnt do
   for j:=1 to 4 do map[i,j]:=0;
  for i:=0 to a1 do
   for j:=0 to c1 do
    for k:=0 to g1 do
     for x:=0 to t1 do hash[i,j,k,x]:=0;

  cnt:=1;
  for i:=1 to n do
  begin
   readln(ch);
   len:=length(ch);
   for j:=1 to len do
   begin
    case ch[j] of
     'A':a[j]:=1;
     'C':a[j]:=2;
     'G':a[j]:=3;
     'T':a[j]:=4;
    end;
   end;
   build;
  end;
  acauto;
  readln(ch);
  len:=length(ch); a1:=0; c1:=0; g1:=0; t1:=0;
  for i:=1 to len do
  begin
   case ch[i] of
    'A':inc(a1);
    'C':inc(c1);
    'G':inc(g1);
    'T':inc(t1);
   end;
  end;
  num:=0;
  for i:=0 to a1 do
   for j:=0 to c1 do
    for k:=0 to g1 do
     for x:=0 to t1 do
     begin
      inc(num); hash[i,j,k,x]:=num;
     end;

 for i:=1 to cnt do
  for j:=0 to a1 do
   for k:=0 to c1 do
    for x:=0 to g1 do
     for y:=0 to t1 do dp[i,hash[j,k,x,y]]:=-1;
  dp[1,hash[0,0,0,0]]:=0;

  ans:=0;
  for i:=0 to a1 do
   for j:=0 to c1 do
    for k:=0 to g1 do
     for x:=0 to c1 do
     begin
      if i+j+k+x=0 then continue;
      s1:=hash[i,j,k,x];
      for u:=1 to cnt do
       for t:=1 to 4 do
       begin
        v:=map[u,t];
        s2:=0;
        if (t=1)and(i>=1) then s2:=hash[i-1,j,k,x]
        else if (t=2)and(j>=1) then s2:=hash[i,j-1,k,x]
         else if (t=3)and(k>=1) then s2:=hash[i,j,k-1,x]
          else  if (t=4)and(x>=1) then s2:=hash[i,j,k,x-1];
        if s2=0 then continue;
        if dp[u,s2]=-1 then continue;
        dp[v,s1]:=max(dp[v,s1],dp[u,s2]+size[v]);
        ans:=max(ans,dp[v,s1]);
       end;
     end;
  writeln('Case ',cas,': ',ans);
 end;
 close(input);
 close(output);
end.

 UPD（2018.10.27）：C++重写 果然C++常数小

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef vector<int> VI;
#define fi first
#define se second
#define MP make_pair
#define N   610
#define M   7010
#define eps 1e-8
#define pi  acos(-1)
#define oo  1e9
#define MOD 10007

int nxt[N][4],q[N],size[N],fa[N],num[41][41][41][41],dp[N][16010],
    A,C,G,T,len,cnt;
char b[M],ch[M];

void build()
{
    int u=1;
    for(int i=1;i<=len;i++)
    {
        int t;
        if(b[i]=='A') t=0;
        if(b[i]=='C') t=1;
        if(b[i]=='G') t=2;
        if(b[i]=='T') t=3;
        if(!nxt[u][t]) nxt[u][t]=++cnt;
        u=nxt[u][t];
    }
    size[u]++;
}
            
void acauto()
{
    int t=0; int w=1; q[1]=1;
    while(t<w)
    {
        int u=q[++t];
        size[u]+=size[fa[u]];
        for(int i=0;i<=3;i++)
        {
             if(nxt[u][i])
              {
                 int son=nxt[u][i];        
                 int p=fa[u];
                 if(u==1) fa[son]=1;
                  else fa[son]=nxt[p][i];
                 q[++w]=son;
             }
              else
              {
                 int p=fa[u];
                 if(u==1) nxt[u][i]=1;
                  else nxt[u][i]=nxt[p][i];
              }
        }
    }
}



int main()
{
    //freopen("hdoj3341.in","r",stdin);
    //freopen("hdoj3341.out","w",stdout);
    int n;
    int cas=0;
    A=C=G=T=cnt=0;
    while(scanf("%d",&n)!=EOF) 
    {
        if(n==0) break;
        cas++;
        for(int i=1;i<=cnt;i++) size[i]=fa[i]=0;
        for(int i=1;i<=cnt;i++)
         for(int j=0;j<=3;j++) nxt[i][j]=0;
        for(int i=0;i<=A;i++)
         for(int j=0;j<=C;j++)
          for(int k=0;k<=G;k++)
           for(int x=0;x<=T;x++) num[i][j][k][x]=0; 
        cnt=1;
        for(int i=1;i<=n;i++){scanf("%s",b+1); len=strlen(b+1); build();}
        acauto();
        scanf("%s",ch+1);
        len=strlen(ch+1);
        A=C=G=T=0;
        for(int i=1;i<=len;i++)
        {
            if(ch[i]=='A') A++;
            if(ch[i]=='C') C++;
            if(ch[i]=='G') G++;
            if(ch[i]=='T') T++;
        }
        int s=0;
        for(int i=0;i<=A;i++)
         for(int j=0;j<=C;j++)
          for(int k=0;k<=G;k++)
           for(int x=0;x<=T;x++) num[i][j][k][x]=++s;
        for(int i=1;i<=cnt;i++)
         for(int j=0;j<=A;j++)
          for(int k=0;k<=C;k++)
           for(int x=0;x<=G;x++)
            for(int y=0;y<=T;y++) dp[i][num[j][k][x][y]]=-1;
        dp[1][num[0][0][0][0]]=0;   
        int ans=0;
        for(int i=0;i<=A;i++)
         for(int j=0;j<=C;j++)
          for(int k=0;k<=G;k++)
           for(int x=0;x<=T;x++)
           {
                    if(i+j+k+x==0) continue;
                    int s1=num[i][j][k][x];
                    for(int u=1;u<=cnt;u++)
                 for(int t=0;t<=3;t++)
                   {
                       int v=nxt[u][t];
                       int s2=0;
                       if(t==0&&i>=1) s2=num[i-1][j][k][x];
                       if(t==1&&j>=1) s2=num[i][j-1][k][x];
                    if(t==2&&k>=1) s2=num[i][j][k-1][x];
                    if(t==3&&x>=1) s2=num[i][j][k][x-1];
                    if(s2==0||dp[u][s2]==-1) continue;
                    dp[v][s1]=max(dp[v][s1],dp[u][s2]+size[v]);
                    ans=max(ans,dp[v][s1]);
                 }
            }
        printf("Case %d: %d
",cas,ans);   
    }
    return 0;
}