最短子序列(最短摘要)

题目：有一段广告，它由很多单词构成，然后它有个摘要，即关键字，问包含摘要(关键字)的在广告正文中的最短字符串是什么？
         简而言之，就是说求在原字符串中包含所有关键字的最短子序列。

算法思想:
             将每个关键字在原字符串中位置记录在一个数组中，每个关键字对应一个数组，然后令distance = max(a[i], b[j], c[k]) - min(a[i], b[j], c[k]),然后最小的那个元素被它所在数组的下一个元素取代，依次循环下去。这里假设是三个关键字，然后就有三个数组。

#include <iostream>
using namespace std;

int max(int a, int b, int c);
int min(int a, int b, int c);

int main()
{
    const int LEN = 10;
    const int MAX = 100000000;
    char** str = new char*[LEN];
    str[0] = "My";
    str[1] = "name";
    str[2] = "is";
    str[3] = "wang";
    str[4] = "name";
    str[5] = "shuai";
    str[6] = "your";
    str[7] = "name";
    str[8] = "is";
    str[9] = "wang";

    char** keyword  =new char*[3];
    keyword[0] = "wang";
    keyword[1] = "name";
    keyword[2] = "your";
    int a[LEN];
    int b[LEN];
    int c[LEN];
    for(int i = 0; i < LEN; i++)
        a[i] = b[i] = c[i] = -1;
    int j = 0;
    int k = 0;
    int l = 0;

    for(int m = 0; m < 3; m++)
    {
        for(int i = 0; i < LEN; i++)
        {
            if(strcmp(str[i], keyword[m]) == 0 && m == 0)
                a[j++] = i;
            else if(strcmp(str[i], keyword[m]) == 0 && m == 1)
                b[k++] = i;
            else if(strcmp(str[i], keyword[m]) == 0 && m == 2)
                c[l++] = i;
        }
    }
    //相隔的最短距离
    int distance;
    //三个关键字的位置
    int postion1;
    int postion2;
    int postion3;

    //用于循环的三个变量
    int p = 0;
    int q = 0;
    int r = 0;


    int minium = MAX;
    while(a[p] >= 0 && b[q] >= 0 && c[r] >= 0)
    {
        distance = max(a[p], b[q], c[r]) - min(a[p], b[q], c[r]);
        if(distance < minium)
        {
            minium = distance;
            postion1 = a[p];
            postion2 = b[q];
            postion3 = c[r];
        }
        if(min(a[p], b[q], c[r]) == a[p])
            p++;
        else if(min(a[p], b[q], c[r]) == b[q])
            q++;
        else if(min(a[p], b[q], c[r]) == c[r])
            r++;
    }

    if(minium != MAX)
    {
        cout<<"长度为："<<minium + 1<<endl;
        cout<<"从位置"<<min(postion1, postion2, postion3) + 1<<"到位置"<<max(postion1, postion2, postion3) + 1<<endl;
    }
    else
        cout<<"NO"<<endl;
    return 0;
}

int max(int a, int b, int c)
{
    int max = a > b ? a : b;
    if(c > max)
        max = c;
    return max;
}

int min(int a, int b, int c)
{
    int min = a < b ? a : b;
    if(c < min)
        min = c;
    return min;
}