ZOJ 3643 Keep Deleting【KMP+栈模拟】

Assume that string A is the substring of string B if and only if we can find A in B, now we have a string A and another string B, your task is to find a A in B from B's left side to B's right side, and delete it from B till A is not a substring of B, then output the number of times you do the delete.

There are only letters(A-Z, a-z) in string A and B.

Input

This problem contains multiple test cases. Each case contains two line, the first line is string A, the second line is string B, the length of A is less than 256, the length of B is less than 512000.

Output

Print exactly one line with the number of times you do the delete for each test case.

Sample Input

abcd
abcabcddabcdababcdcd

Sample Output

5

Hint

abcabcddabcdababcdcd delete=0
abcdabcdababcdcd     delete=1
abcdababcdcd         delete=2
ababcdcd             delete=3
abcd                 delete=4
                     delete=5

题目大意：给出a和b串，a是b串的子串，如果b串有连续的a串，那么就将b串的a串删除，问删除多少次；
思路：栈的模拟思想，每个字符入栈，如果出现和串a相同的子串那么相同的字串退栈；剩下的字符继续入栈；但是暴力的找很消耗时间，因此用到KMP;
　　例如 串a:abcd 串b:abcabcdd
　　在b串中找到了a串，但是应该跳道德是abc串的c那里看下一个字符是否是d;所以先用KMP的=求出串a的next[]数组的值；

代码如下：

栈写的代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stack>
using namespace std;
#define N 512005
char cha[N], chb[N];
int next[N];
int main()
{
    int i, j, lena, lenb, t, ans;
    while(scanf("%s%s", cha+1, chb+1)!=EOF)
    {
        lena=strlen(cha+1), lenb=strlen(chb+1);
        memset(next, 0, sizeof(next));
        next[1]=0, j=0;
        stack<int>s;
        for(i=2; i<=lena; i++)
        {
            while(j&&cha[j+1]!=cha[i])
                j=next[j];
            if(cha[j+1]==cha[i])
                j++;
            next[i]=j;
        }
        j=0, ans=0;
        for(i=1; i<=lenb; i++)
        {
            while(j&&cha[j+1]!=chb[i])
                j=next[j];
            if(cha[j+1]==chb[i])
                j++;
            s.push(j);
            if(j==lena)
            {
                ans++;
                for(t=0;t<lena;t++)
                    s.pop();
                if(!s.empty())
                    j=s.top();
                else
                    j=0;
            }
        }
        printf("%d\n", ans);
    }
}

数组模拟栈代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define N 512005
char cha[N], chb[N];
int next[N], stack[N];
int main()
{
    int i, j, lena, lenb, t, ans, tt;
    while(scanf("%s%s", cha+1, chb+1)!=EOF)
    {
        lena=strlen(cha+1), lenb=strlen(chb+1);
        memset(next, 0, sizeof(next));
        memset(stack, 0, sizeof(stack));
        next[1]=0, j=0;
        for(i=2; i<=lena; i++)
        {
            while(j&&cha[j+1]!=cha[i])
                j=next[j];
            if(cha[j+1]==cha[i])
                j++;
            next[i]=j;
        }
        j=0, ans=0, tt=0;
        for(i=1; i<=lenb; i++)
        {
            while(j&&cha[j+1]!=chb[i])
                j=next[j];
            if(cha[j+1]==chb[i])
                j++;
            stack[tt++]=j;
            if(j==lena)
            {
                ans++;
                for(t=0;t<lena;t++)
                    tt--;
                if(tt==0)
                    j=0;
                else
                    j=stack[tt-1];
            }
        }
        printf("%d\n", ans);
    }
}