地址：http://acm.split.hdu.edu.cn/showproblem.php?pid=4758
题目：
Walk Through Squares
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1548 Accepted Submission(s): 514
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.
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
3 2
RRD
DDR
3 2
R
D
Sample Output
1
10
10
Source
思路：
明显ac自动机+dp。
dp[i][x][y][s]：表示走了i步，到达(x,y)位置后，状态为s的方案数。（s是包含目标串状态的压缩）
这样的dp比较浪费空间，因为y可以通过i-x推出，所以dp状态应该是：dp[i][x][s]。
这题还要你滚动数组。。。
#include <queue>
#include <cstring>
#include <cstdio>
using namespace std; const int mod = 1e9 + ;
struct AC_auto
{
const static int LetterSize = ;
const static int TrieSize = * ( 4e2 + ); int tot,root,fail[TrieSize],end[TrieSize],next[TrieSize][LetterSize];
int dp[][TrieSize][][]; int newnode(void)
{
memset(next[tot],-,sizeof(next[tot]));
end[tot] = ;
return tot++;
} void init(void)
{
tot = ;
root = newnode();
} int getidx(char x)
{
return x=='R';
} void insert(char *ss,int id)
{
int len = strlen(ss);
int now = root;
for(int i = ; i < len; i++)
{
int idx = getidx(ss[i]);
if(next[now][idx] == -)
next[now][idx] = newnode();
now = next[now][idx];
}
end[now]|=id;
} void build(void)
{
queue<int>Q;
fail[root] = root;
for(int i = ; i < LetterSize; i++)
if(next[root][i] == -)
next[root][i] = root;
else
fail[next[root][i]] = root,Q.push(next[root][i]);
while(Q.size())
{
int now = Q.front();Q.pop();
for(int i = ; i < LetterSize; 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]),end[next[now][i]]|=end[next[fail[now]][i]];
}
} int match(char *ss)
{
int len,now,res;
len = strlen(ss),now = root,res = ;
for(int i = ; i < len; i++)
{
int idx = getidx(ss[i]);
int tmp = now = next[now][idx];
while(tmp)
{
res += end[tmp];
end[tmp] = ;//按题目修改
tmp = fail[tmp];
}
}
return res;
} void go(int n,int m)
{
//debug();
int now=;
memset(dp[],,sizeof dp[]);
dp[][][][]=;
for(int p=;p<n+m;p++)
{
memset(dp[now],,sizeof dp[now]);
for(int i=;i<tot;i++)
for(int x=;x<=n;x++)
for(int k=;k<&&x<=p&&p-x<=m;k++)
if(dp[now^][i][x][k])
{
if(x!=n)
{
int nt=next[i][],st=end[nt]|k;
dp[now][nt][x+][st] = (dp[now][nt][x+][st] + dp[now^][i][x][k] ) % mod;
}
if(p-x!=m)
{
int nt=next[i][],st=end[nt]|k;
dp[now][nt][x][st] = (dp[now][nt][x][st] + dp[now^][i][x][k] ) % mod;
}
}
// printf("=======\n");
// for(int i=0;i<tot;i++)
// for(int x=0;x<=n&&x<=p+1&&p+1-x<=m;x++)
// for(int k=0;k<4;k++)
// printf("%d %d %d %d :%d\n",now,i,x,k,dp[now][i][x][k]);
now^=;
}
int ans=;
for(int i=;i<tot;i++)
ans=(ans+dp[now^][i][n][])%mod;
printf("%d\n",ans);
}
void debug()
{
for(int i = ;i < tot;i++)
{
printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
for(int j = ;j < LetterSize;j++)
printf("%3d",next[i][j]);
printf("]\n");
}
}
}ac;
char ss[];
int main(void)
{
int t,n,m;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%s",&m,&n,ss);
ac.init();
ac.insert(ss,);
scanf("%s",ss);
ac.insert(ss,);
ac.build();
ac.go(n,m);
}
return ;
}