Letter Grid（动规）

Consider the following letter grid:

E

R

A

T

A

T

S

R

A

U

T

U

There are 7 ways to read the word TARTU from the grid:

E

R

A

T

A

T

S

R

A

U

T

U

 

E

R

A

T

A

T

S

R

A

U

T

U

 

E

R

A

T

A

T

S

R

A

U

T

U

 

E

R

A

T

A

T

S

R

A

U

T

U

 

E

R

A

T

A

T

S

R

A

U

T

U

 

E

R

A

T

A

T

S

R

A

U

T

U

 

E

R

A

T

A

T

S

R

A

U

T

U

 

 

Given a letter grid and a word, your task is to determine the number of ways the word can be read from the grid. The first letter of the word can be in any cell of the grid, and after each letter, the next letter has to be in one of the neighbour cells (horizontally, vertically or diagonally). A cell can be used multiple times when reading the word.

Input Data

Each test case contains three integers: H (1 ≤ H ≤ 200), the height of the grid, W (1 ≤ W ≤ 200), the width of the grid, and L (1 ≤ L ≤ 100), the length of the word. The following H lines each containing W letters describe the grid. The last line containing L letters describes the word. All letters in the grid and in the word are uppercase English letters (A...Z).

Output Data

For each case, output only line contained one integer: the number of ways the word can be read from the grid. You may assume that the answer is always at most 10¹⁸.

Examples

in

out

3 4 5
ERAT
ATSR
AUTU
TARTU

2 2 10
AA
AA
AAAAAAAAAA

7

78732

 

 这是一道动态规划的题：

即待匹配的字符串，每一个字符依次代表一种状态，最后一个字符即代表最后的状态。

eg.

    第一个例子：

 TARTU

T:                  A:                 R:                     T:                    U:

0 0 0 1           0 0 2 0          0 3 0 0              0 0 0 2             0 0 0 0

0 1 0 0           1 0 0 0          0 0 0 2              0 3 0 0             0 0 0 0

0 0 1 0           0 0 0 0          0 0 0 0              0 0 2 0             0 5 0 2

其中每一个不为零的位置，代表到达对应字符的方法数。

后一种字符的方法数，为其前一种字符的得方法数在相应位置的叠加。

 

代码实现：

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;char a[201][201];long long sum[201][201][101];char ex[101];int strx[8]={0,0,-1,1,1,-1,1,-1};int stry[8]={1,-1,0,0,1,-1,-1,1};int main(){    int n,m,le;    while(scanf("%d%d%d",&n,&m,&le)!=EOF){        memset(sum,0,sizeof(sum));        int times=0;        for(int i=0;i<n;i++){                scanf("%s",a[i]);                        }        scanf("%s",ex);        for(int j=0;j<n;j++){                for(int k=0;k<m;k++){                        if(a[j][k]==ex[0]){                                times++;                                sum[j][k][0]=1;                                }                                }                                }        if(le==1){              cout<<times<<endl;              continue;              }        long long resum=0;        for(int g=1;g<le;g++){                long long res;                for(int f=0;f<n;f++){                        for(int d=0;d<m;d++){                                if(a[f][d]==ex[g]){                                      res=0;                                      for(int s=0;s<8;s++){                                              int temp1=f+strx[s];                                              int temp2=d+stry[s];                                              if(temp1>=0&&temp1<n&&temp2>=0&&temp2<m){                                                       if(a[temp1][temp2]==ex[g-1]){                                                              res+=sum[temp1][temp2][g-1];                                                              }                                                              }                                                              }                                      sum[f][d][g]=res;                                      if(g==le-1)                                          resum+=res;                                      }                                      }                                      }                                      }        cout<<resum<<endl;        }    return 0;}