hdu 4758 Walk Through Squares(AC自动机+DP，4级)

Walk Through Squares

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 234    Accepted Submission(s): 58

Problem Description

  On the beaming day of 60th anniversary of NJUST, as a military college which was Second Artillery Academy of Harbin Military Engineering Institute before, queue phalanx is a special landscape.
  
  Here is a M*N rectangle, and this one can be divided into M*N squares which are of the same size. As shown in the figure below:
  01--02--03--04
  || || || ||
  05--06--07--08
  || || || ||
  09--10--11--12
  Consequently, we have (M+1)*(N+1) nodes, which are all connected to their adjacent nodes. And actual queue phalanx will go along the edges.
  The ID of the first node,the one in top-left corner,is 1. And the ID increases line by line first ,and then by column in turn ,as shown in the figure above.
  For every node,there are two viable paths:
  (1)go downward, indicated by 'D';
  (2)go right, indicated by 'R';
  The current mission is that, each queue phalanx has to walk from the left-top node No.1 to the right-bottom node whose id is (M+1)*(N+1).
In order to make a more aesthetic marching, each queue phalanx has to conduct two necessary actions. Let's define the action:
  An action is started from a node to go for a secified tpravel mode.

  So, two actions must show up in the way from 1 to (M+1)*(N+1).

  For example, as to a 3*2 rectangle, figure below:
    01--02--03--04
    || || || ||
    05--06--07--08
    || || || ||
    09--10--11--12
  Assume that the two actions are (1)RRD (2)DDR

  As a result , there is only one way : RRDDR. Briefly, you can not find another sequence containing these two strings at the same time.
  If given the N, M and two actions, can you calculate the total ways of walking from node No.1 to the right-bottom node ?

 

Input

  The first line contains a number T,(T is about 100, including 90 small test cases and 10 large ones) denoting the number of the test cases.
  For each test cases,the first line contains two positive integers M and N(For large test cases,1<=M,N<=100, and for small ones 1<=M,N<=40). M denotes the row number and N denotes the column number.
  The next two lines each contains a string which contains only 'R' and 'D'. The length of string will not exceed 100. We ensure there are no empty strings and the two strings are different.

 

Output

  For each test cases,print the answer MOD 1000000007 in one line.

 

Sample Input

   
   
   
   

    
    
    
    2
3 2
RRD
DDR
3 2
R
D
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
10
   
   
   
   

 

Source

2013 ACM/ICPC Asia Regional Nanjing Online

 

Recommend

liuyiding

 

思路：AC自动机用上失败指针，预处理出转移态。然后DP[x][y][statu][con] x，y点statu 态，包含子串状态为con 的最优值。

        哎，感觉就是AC自动+DP的模板题，居然当时没YY出来，真是失败啊。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mod=1000000007;
const int mm=109;
const int msize=209;
class AC_Machine
{
  public:
    int ch[msize][2],val[msize],f[msize],sz;
    void clear()
    {
      sz=1;clr(ch[0],0),val[0]=0;
    }
    int idx(char z)
    {
      if(z=='D')return 0;
      if(z=='R')return 1;
    }
    void insert(char*s,int cc)
    {
      int u=0,v,c;
      for(int i=0;s[i];++i)
      {
        c=idx(s[i]);
        if(!ch[u][c])
        {
          clr(ch[sz],0);val[sz]=0;
          ch[u][c]=sz++;
        }
        u=ch[u][c];
      }
      val[u]|=cc;
    }
    void getFail()
    {
      int u,v,r;
      queue<int>Q;
      FOR(i,0,1)
      {
        u=ch[0][i];
        if(u){Q.push(u);f[u]=0; }
      }
      while(!Q.empty())
      {
        r=Q.front();Q.pop();
        val[ r ]|=val[ f[r] ];
        FOR(c,0,1)
        {
          u=ch[r][c];
          if(!u){ ch[r][c]=ch[ f[r] ][c];continue; }
          v=f[r];Q.push(u);
          f[u]=ch[v][c];
        }
      }
    }
   int dp[mm][mm][msize][4];
   void solve(int n,int m)
   { int tmp,now,then;
     clr(dp,0);
     dp[0][0][ val[0] ][0]=1;
     FOR(x,0,n)FOR(y,0,m)
     FOR(k,0,sz-1)FOR(l,0,3)
     {
       if(x+1<=n)///down
       {
         then=ch[k][0];tmp=l|val[then];
         dp[x+1][y][then][tmp]=(dp[x+1][y][then][tmp]+dp[x][y][k][l])%mod;
       }
       if(y+1<=m)
       {
         then=ch[k][1];tmp=l|val[then];
         dp[x][y+1][then][tmp]=(dp[x][y+1][then][tmp]+dp[x][y][k][l])%mod;
       }
     }
     int ans=0;
     FOR(i,0,sz-1)//FOR(j,0,3)两种动作都要有
     {
       ans=(ans+dp[n][m][i][3])%mod;
     }
     printf("%d\n",ans);
   }
};
AC_Machine ac;
char s[2][msize];
int main()
{
  int cas;
   scanf("%d",&cas);

    while(cas--)
    { int n,m;
      scanf("%d%d",&m,&n);
      ac.clear();
      scanf("%s%s",s[0],s[1]);
      ac.insert(s[0],1);
      ac.insert(s[1],2);
      ac.getFail();
      ac.solve(n,m);
    }

}