hdu4758AC自动机+状态压缩DP

http://acm.hdu.edu.cn/showproblem.php?pid=4758

  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 ?

2
3 2
RRD
DDR
3 2
R
D

1
10

/**
hdu 4758 AC自动机+状态压缩DP
题目大意：给定一个n*m的棋盘，要求从左上角走到右下角，每次只能向下或者向右走。给定两个模式串，问有多少种走法
          满足每一种走法含有这两个模式串作为子串
解题思路：构造自动机，状态压缩包含的子串。
          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。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <iostream>
using namespace std;
const int mod=1e9+7;
int dp[110][110][220][4];
int n,m;
struct Trie
{
    int next[420][2],fail[420],end[420];
    int root,L;
    int change(char ch)
    {
        if(ch=='R')return 0;
        return 1;
    }
    int newnode()
    {
        for(int i=0; i<2; i++)
        {
            next[L][i]=-1;
        }
        end[L++]=0;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    void insert(char *buf,int id)
    {
        int len=strlen(buf);
        int now=root;
        for(int i=0; i<len; i++)
        {
            if(next[now][change(buf[i])]==-1)
                next[now][change(buf[i])]=newnode();
            now=next[now][change(buf[i])];
        }
        end[now]|=(1<<id);
    }
    void build()
    {
        queue<int>Q;
        fail[root]=root;
        for(int i=0; i<2; i++)
        {
            if(next[root][i]==-1)
                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=0; i<2; i++)
            {
                if(next[now][i]==-1)
                {
                    next[now][i]=next[fail[now]][i];
                }
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    int solve()
    {
        memset(dp,0,sizeof(dp));
        dp[0][0][0][0]=1;
        for(int x=0; x<=n; x++)
        {
            for(int y=0; y<=m; y++)
            {
                // printf("**\n");
                for(int i=0; i<L; i++)
                {
                    for(int k=0; k<4; k++)
                    {
                        if(dp[x][y][i][k]==0)continue;
                        if(x<n)
                        {
                            int nxt=next[i][0];
                            dp[x+1][y][nxt][k|end[nxt]]+=dp[x][y][i][k];
                            dp[x+1][y][nxt][k|end[nxt]]%=mod;
                        }
                        if(y<m)
                        {
                            int nxt=next[i][1];
                            dp[x][y+1][nxt][k|end[nxt]]+=dp[x][y][i][k];
                            dp[x][y+1][nxt][k|end[nxt]]%=mod;
                        }
                    }
                }
            }
        }
        int ret=0;
        for(int i=0; i<L; i++)
        {
            ret+=dp[n][m][i][3];
            ret%=mod;
        }
        return ret;
    }
} ac;
char str[210];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        ac.init();
        for(int i=0; i<2; i++)
        {
            scanf("%s",str);
            ac.insert(str,i);
        }
        ac.build();
        printf("%d\n",ac.solve());
    }
    return 0;
}

hdu4758AC自动机+状态压缩DP

原文：http://blog.csdn.net/lvshubao1314/article/details/45476405