Given a string S of length not exceeding 10 ^ 5, and set A with n elements(n≤100) consisting of ASCII characters, are case sensitive. Look for consecutive substring in S that contain the complete set of characters in A and the shortest length.
(Consecutive substring is the string with the characters in the order of A and can be far apart)
Data on: file GETSTRING.INP
The first line contains contains a string S
The first line 2 the number n.
The next N lines, each containing 1 A character in the set.
Data out: file GETSTRING.OUT
Including 1 single line recorded the shortest length find substring.
Example:
GETSTRING.INP
hello the gioin
4
t
and
g
the
GETSTRING.OUT
7
———————
call F[in][j] is the distance from the characters A[in] (The[in] = s[j]) to A[i-1] in string S.
I need to find the string length is the shortest distance from A[n] to A[1].
F[in][j] is calculated as follows:
Browse array A, with each element of the array A string s approval if A[in] = s[j] then we perform:
if the A[0] then f[in][j] = 1;
reverse f[in][j] = j-start + temp;
in which the start is located near the most j f[i-1][start] > 0 and temp is the value f[i-1][start] (length from S[start] to S[x] that S[x] =A[1])
Finally KQ is of f min[A-1][j] (j=0 -> ns, with min > 0, if no values >0 then min = 0)

The test results and F[in][j] on ideone.com

#include<cstdio>
#include<cstdlib>
#include <string>
#include <iostream>
using namespace std;
int main()
{
        string s;
        getline(cin,s);
        int n;
        scanf("%d",&n);
        char A[n];
        int f[n][s.length()];
        
        
        for (int i=0; i<n; i++)
        {
                scanf("%c",&A[i]);
                scanf("%c",&A[i]);
                //printf("%c  ",A[i]);
        }
        //cout<<endl<<s<<endl;
        //printf("%dn",n);
        for (int i=0; i<n; i++)
                for (int j=0; j<s.length(); j++)
                        f[i][j] = 0;
        
        int begin = 0, min = 0;
        for (int i=0; i<n; i++)
        {
                int temp = 0, check = 0, start = 0;  
                //temp la gia tri do dai tai no, check: neu co 1 lan bang roi, start noi bat dau xet
                for (int j=begin; j<s.length(); j++)
                {
                        if (i>0 && f[i-1][j] > 0) { start = j; temp = f[i-1][j]; } // tim start va temp
                        if (A[i] == s[j])
                        {
                                //printf("%3c%3d%3dn",A[i],i,j);
                                if (check == 0) begin = j;      // chua co ky tu nao bang thi j la noi bat dau
                                if (i==0) f[i][j] = 1;          // xet cho A[0]
                                else
                                {
                                        f[i][j] = j-start + temp; //vi tri hien tai - noi bat dau + gia tri do dai
                                }
                                check = 1;
                                if (i==n-1)
                                {
                                        if (f[i][j] < min || min == 0) min = f[i][j];
                                }
                        }
                }
        }
        for (int i=0; i<n; i++)
        {
                for (int j=0; j<s.length(); j++)
                        printf("%4d",f[i][j]);
                printf("n");
        }
        printf("%d",min);       
        return 0;
}

One other solutions refer here