HDU 4758 Walk Through Squares （2013南京网络赛1011题，AC自动机+DP）

Walk Through Squares

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 200    Accepted Submission(s): 57

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 specified travel 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

 

太伤了。

网络赛的时候没有过这题，都写对了，就是一直WA

比赛结束后问了下世界冠军cxlove,看了一眼发现是建立AC自动机的时候，fail的end要传递下，加一句话就过了，TAT

使用AC自动机+DP。

dp[x][y][i][k] 四维DP，表示R的个数是x,D的个数是y, i是在AC自动机上的节点编号。k 是状态压缩。

 /* ***********************************************
 Author        :kuangbin
 Created Time  :2013/9/21 星期六 15:57:51
 File Name     :2013南京网络赛\1011.cpp
 ************************************************ */

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <stdio.h>
 #include <string.h>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;
 const int MOD = 1e9+;

 int dp[][][][];
 int n,m;
 //struct Trie
 //{
     int next[][],fail[],end[];
     int root,L;
     inline int change(char ch)
     {
         if(ch == 'R')return ;
         else return ;
     }
     inline int newnode()
     {
         for(int i = ;i < ;i++)
             next[L][i] = -;
         end[L++] = ;
         return L-;
     }
     inline void init()
     {
         L = ;
         root = newnode();
     }
     inline void insert(char buf[],int id)
     {
         int len = strlen(buf);
         int now = root;
         for(int i = ;i < len;i++)
         {
             if(next[now][change(buf[i])] == -)
                 next[now][change(buf[i])] = newnode();
             now = next[now][change(buf[i])];
         }
         end[now] |= (<<id);
     }
     inline void build()
     {
         queue<int>Q;
         fail[root] = root;
         for(int i = ;i < ;i++)
             if(next[root][i] == -)
                 next[root][i] = root;
             else
             {
                 fail[next[root][i]] = root;
                 Q.push(next[root][i]);
             }
         while( !Q.empty() )
         {
             int now = Q.front();
             Q.pop();
             end[now] |= end[fail[now]];
             for(int i = ;i < ;i++)
                 if(next[now][i] == -)
                     next[now][i] = next[fail[now]][i];
                 else
                 {
                     fail[next[now][i]]=next[fail[now]][i];
                     Q.push(next[now][i]);
                 }
         }
     }
     inline int solve()
     {
         dp[][][][] = ;
         int ret = ;

             for(int x = ;x <= n;x++)
                 for(int y = ;y <= m;y++)
                     for(int i = ;i < L;i++)

                     for(int k = ;k < ;k++)

                     {
                         if(dp[x][y][i][k] == )continue;
                         if(x < n)
                         {
                             int nxt = next[i][];
                             dp[x+][y][nxt][k|end[nxt]] += dp[x][y][i][k];
                             if(dp[x+][y][nxt][k|end[nxt]] >= MOD)
                                 dp[x+][y][nxt][k|end[nxt]] -= MOD;
                         }
                         if(y < m)
                         {
                             int nxt = next[i][];
                             dp[x][y+][nxt][k|end[nxt]] += dp[x][y][i][k];
                             if(dp[x][y+][nxt][k|end[nxt]] >= MOD)
                                 dp[x][y+][nxt][k|end[nxt]] -= MOD;
                         }
                     }
         for(int i = ;i < L;i++)
         {
             ret += dp[n][m][i][];
             if(ret >= MOD)ret -= MOD;
         }

         return ret;

     }
 //};
 //Trie ac;
 char str[];

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int T;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&m);
         init();
         for(int i = ;i < ;i++)
         {
             scanf("%s",str);
             insert(str,i);
         }
         build();
         for(int i = ;i <= n;i++)
             for(int j = ;j <= m;j++)
                 for(int x = ; x < L;x++)
                     for(int y = ;y < ;y++)
                         dp[i][j][x][y] = ;
         printf("%d\n",solve());
     }
     return ;
 }