POJ 3852 StringLD 左字符串模拟

最新推荐文章于 2021-04-05 15:05:50 发布

最新推荐文章于 2021-04-05 15:05:50 发布 · 102 阅读

本文介绍了一个字符串处理问题，要求在给定的单词列表中，连续从每个单词左侧删除字符直至出现空字符串或重复字符串，旨在寻找最多能进行几步操作。通过逆序存储字符串并记录长度信息来提高效率。

String LD

Time Limit:1000MS		Memory Limit:65536K

Description

Stringld (left delete) is a function that gets a string and deletes its leftmost character (for instance Stringld(“acm”) returns “cm”).

You are given a list of distinct words, and at each step, we apply stringld on every word in the list. Write a program that determines the number of steps that can be applied until at least one of the conditions become true:

1) A word becomes empty string, or
2) a duplicate word is generated.

For example, having the list of words aab, abac, and caac, applying the function on the input for the first time results in ab, bac, and aac. For the second time, we get b, ac, and ac. Since in the second step, we have two ac strings, the condition 2 is true, and the output of your program should be 1. Note that we do not count the last step that has resulted in duplicate string. More examples are found in the sample input and output section.

Input

There are multiple test cases in the input. The first line of each test case is n (1 <= n <= 100), the number of words.
Each of the next n lines contains a string of at most 100 lower case characters.
The input terminates with a line containing 0.

Output

For each test case, write a single line containing the maximum number of stringld we can call.

Sample Input

4
aaba
aaca
baabcd
dcba
3
aaa
bbbb
ccccc
0

Sample Output

方法点击，由于对字符串从左边删除字符时间复杂度比较大，虽然本题的数据规模不是很大，可以在存储的时候将字符串逆置，然后再进行字符串操作，这里为了减少时间复杂度，额外使用了数组来保存记录字符串的长度信息。

代码如下：

/**************************************************************************
* Problem: POJ 3852 StringLD
* Copyright 2011 by Yan
* DATE:
* E-Mail: yming0221@gmail.com
************************************************************************/
#include <stdio.h>

int n;
char value[101][101];
int len[101];/*保存字符串的长度*/
char cache[101];

int main()
{
    int i,j,k,cnt;
    /*freopen("input","r",stdin);*/
    while(scanf("%d",&n)&&n!=0)
    {
    	cnt=0;
    	for(i=0;i<n;i++) len[i]=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",cache);
            len[i]=strlen(cache);
            for(j=0;j<len[i];j++)/*实现字符串的逆序存储*/
			{
				value[i][j]=cache[len[i]-j-1];
			}
			value[i][len[i]]='\0';
        }
        while(1)
        {
        	cnt++;
			for(i=0;i<n;i++)/*每个字符串删除一个字符*/
			{

				len[i]--;
				value[i][len[i]]='\0';
			}
			for(i=0;i<n;i++)
			{
				if(len[i]==0) goto next;/*有字符串为空串了*/

				for(k=0;k<n;k++)
					for(j=k+1;j<n;j++)
						if(len[k]==len[j])/*减少不必要的字符串比较*/
						{
							/*printf("%s   %s\n",value[k],value[j]);*/
							if(strcmp(value[k],value[j])==0) goto next;
						}
			}
        }
next:

		printf("%d\n",--cnt);
    }
    return 0;
}