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
题意：给你两串，求用m个R。n个D能组成多少个包括这两个串
思路：先构造一个AC自己主动机记录每一个状态包括两个串的状态，然后利用dp[i][j][k][s]表示i个R，j个D。此时AC自己主动机状态位置到k的时候，状态为s时的个数进行转移
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int mod = 1e9+7;
int dp[110][110][220][4];
int n,m;
int nxt[420][2],fail[420],end[420];
int root,cnt;
inline int change(char ch) {
if (ch == 'R')
return 0;
else return 1;
}
inline int newNode() {
for (int i = 0; i < 2; i++)
nxt[cnt][i] = -1;
end[cnt++] = 0;
return cnt-1;
}
inline void init() {
cnt = 0;
root = newNode();
}
inline void insert(char buf[], int id) {
int len = strlen(buf);
int now = root;
for (int i = 0; i < len; i++) {
if (nxt[now][change(buf[i])] == -1)
nxt[now][change(buf[i])] = newNode();
now = nxt[now][change(buf[i])];
}
end[now] |= (1<<id);
}
inline void build() {
queue<int> q;
fail[root] = root;
for (int i = 0; i < 2; i++)
if (nxt[root][i] == -1)
nxt[root][i] = root;
else {
fail[nxt[root][i]] = root;
q.push(nxt[root][i]);
}
while (!q.empty()) {
int now = q.front();
q.pop();
end[now] |= end[fail[now]];
for (int i = 0; i < 2; i++)
if (nxt[now][i] == -1)
nxt[now][i] = nxt[fail[now]][i];
else {
fail[nxt[now][i]] = nxt[fail[now]][i];
q.push(nxt[now][i]);
}
}
}
inline int solve() {
dp[0][0][0][0] = 1;
for (int x = 0; x <= n; x++)
for (int y = 0; y <= m; y++)
for (int i = 0; i < cnt; i++)
for (int k = 0; k < 4; k++) {
if (dp[x][y][i][k] == 0)
continue;
if (x < n) {
int cur = nxt[i][0];
dp[x+1][y][cur][k|end[cur]] += dp[x][y][i][k];
dp[x+1][y][cur][k|end[cur]] %= mod;;
}
if (y < m) {
int cur = nxt[i][1];
dp[x][y+1][cur][k|end[cur]] += dp[x][y][i][k];
dp[x][y+1][cur][k|end[cur]] %= mod;
}
}
int ans = 0;
for (int i = 0; i < cnt; i++) {
ans += dp[n][m][i][3];
ans %= mod;
}
return ans;
}
char str[210];
int main() {
int t;
scanf("%d", &t);
while (t--) {
scanf("%d%d", &n, &m);
init();
for (int i = 0; i < 2; i++) {
scanf("%s", str);
insert(str, i);
}
build();
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++)
for (int x = 0; x < cnt; x++)
for (int y = 0; y < 4; y++)
dp[i][j][x][y] = 0;
printf("%d\n", solve());
}
return 0;
}