hdu 4758 - Walk Through Squares（AC自动机+DP）现场赛

Walk Through Squares

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1046    Accepted Submission(s): 318

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

23 2RRDDDR3 2RD

 

Sample Output

110

 

Source

2013 ACM/ICPC Asia Regional Nanjing Online

题意：包含给定的两个字符串，并且能够从左上角到达右下角，有多少种走法

思路：构建AC自动机后，四维dp[x][y][i][j]表示有x个R，y个D，在树上的节点i，并且状态集为j的方案数，刚开始一直在纠结怎么保证从左上到右下，看了kuangbin大神的题解，恍然大悟

#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<algorithm>using namespace std;typedef long long LL;const int maxn=105*2;const int maxm=105;const int MOD=1e9+7;const double INF=1e20;const int SIGMA_SIZE=2;int N,M;int dp[maxm][maxm][maxn][4];char s[maxm];struct AC{    int ch[maxn][2];    int val[maxn];    int fail[maxn];    int sz;    void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;}    int idx(char x)    {        if(x=='R')return 0;        return 1;    }    void insert(char *s,int id)    {        int u=0;        int n=strlen(s);        for(int i=0;i<n;i++)        {            int c=idx(s[i]);            if(!ch[u][c])            {                memset(ch[sz],0,sizeof(ch[sz]));                val[sz]=0;                ch[u][c]=sz++;            }            u=ch[u][c];        }        val[u]|=(1<<id);    }    void getfail()    {        int u=0;        queue<int> q;        fail[0]=0;        for(int c=0;c<SIGMA_SIZE;c++)        {            u=ch[0][c];            if(u){fail[u]=0;q.push(u);}        }        while(!q.empty())        {            int r=q.front();q.pop();            val[r]|=val[fail[r]];            for(int c=0;c<SIGMA_SIZE;c++)            {                u=ch[r][c];                if(!u){ch[r][c]=ch[fail[r]][c];continue;}                q.push(u);                int v=fail[r];                while(v&&!ch[v][c])v=fail[v];                fail[u]=ch[v][c];            }        }    }    void solve()    {        for(int x=0;x<=N;x++)            for(int y=0;y<=M;y++)                for(int i=0;i<sz;i++)                    for(int j=0;j<4;j++)dp[x][y][i][j]=0;        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<sz;i++)                {                    for(int j=0;j<4;j++)                    {                        if(dp[x][y][i][j]==0)continue;                        if(x<N)                        {                            int v=ch[i][0];                            dp[x+1][y][v][j|val[v]]+=dp[x][y][i][j];                            if(dp[x+1][y][v][j|val[v]]>=MOD)                                dp[x+1][y][v][j|val[v]]-=MOD;                        }                        if(y<M)                        {                            int v=ch[i][1];                            dp[x][y+1][v][j|val[v]]+=dp[x][y][i][j];                            if(dp[x][y+1][v][j|val[v]]>=MOD)                                dp[x][y+1][v][j|val[v]]-=MOD;                        }                    }                }            }        }        int ans=0;        for(int i=0;i<sz;i++)        {            ans+=dp[N][M][i][3];            if(ans>=MOD)ans-=MOD;        }        printf("%d\n",ans);    }}ac;int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&N,&M);        ac.clear();        for(int i=0;i<2;i++)        {            scanf("%s",s);            ac.insert(s,i);        }        ac.getfail();        ac.solve();    }    return 0;}