HDU 4758 Walk Through Squares(AC自动机+DP)

题目链接

难得出一个AC自动机，我还没做到这个题呢。。。这题思路不难想，小小的状压出一维来，不过，D和R，让我wa死了，AC自动机，还得刷啊。。。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define MOD 1000000007
int dp[101][101][201][4];
int trie[201][2];
int o[201];
int fail[201],que[1001],p[201][201];
int t;
void CL()
{
    t = 1;
    memset(trie,-1,sizeof(trie));
    memset(o,0,sizeof(o));
    memset(p,0,sizeof(p));
}
void insert(char *s,int x)
{
    int i,root,len,temp;
    len = strlen(s);
    root = 0;
    for(i = 0; i < len; i ++)
    {
        temp = s[i] == 'D' ? 1:0;
        if(trie[root][temp] == -1)
            trie[root][temp] = t ++;
        root = trie[root][temp];
    }
    o[root] = 1<<x;
}
void build_ac()
{
    int head,tail,front,i;
    head = tail = 0;
    for(i = 0; i < 2; i ++)
    {
        if(trie[0][i] != -1)
        {
            fail[trie[0][i]] = 0;
            que[tail++] = trie[0][i];
        }
        else
            trie[0][i] = 0;
    }
    while(head != tail)
    {
        front = que[head++];
        o[front] |= o[fail[front]];
        for(i = 0; i < 2; i ++)
        {
            if(trie[front][i] != -1)
            {
                que[tail++] = trie[front][i];
                fail[trie[front][i]] = trie[fail[front]][i];
            }
            else
            {
                trie[front][i] = trie[fail[front]][i];
            }
        }
    }
}
int main()
{
    int cas,i,j,k,u,n,m,temp;
    char str[101];
    scanf("%d",&cas);
    while(cas--)
    {
        CL();
        scanf("%d%d",&n,&m);
        for(i = 0; i < 2; i ++)
        {
            scanf("%s",str);
            insert(str,i);
        }
        for(i = 0; i <= n; i ++)
        {
            for(j = 0; j <= m; j ++)
            {
                for(k = 0; k < t; k ++)
                {
                    for(u = 0; u < 4; u ++)
                        dp[i][j][k][u] = 0;
                }
            }
        }
        build_ac();
        dp[0][0][0][0] = 1;
        for(i = 0; i <= n; i ++)
        {
            for(j = 0; j <= m; j ++)
            {
                for(k = 0; k < t; k ++)
                {
                    for(u = 0; u < 4; u ++)
                    {
                        if(i != n)
                        {
                            temp = trie[k][0];
                            dp[i+1][j][temp][u|o[temp]] = (dp[i+1][j][temp][u|o[temp]] + dp[i][j][k][u])%MOD;
                        }
                        if(j != m)
                        {
                            temp = trie[k][1];
                            dp[i][j+1][temp][u|o[temp]] = (dp[i][j+1][temp][u|o[temp]] + dp[i][j][k][u])%MOD;
                        }
                    }
                }
            }
        }
        int ans = 0;
        for(i = 0; i < t; i ++)
        {
            ans = (ans + dp[n][m][i][3])%MOD;
        }
        printf("%d
",ans);
    }
    return 0;
}