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字符串组合判断算法

最新推荐文章于 2020-07-12 21:05:55 发布

原创最新推荐文章于 2020-07-12 21:05:55 发布 · 835 阅读

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本文介绍了一个算法问题，即判断第三个字符串是否能通过前两个字符串的字符按原顺序组合而成。通过对输入字符串进行逐字符对比，利用二维数组记录可能的组合状态，最终得出结论。

Time Limit:

1000ms

Memory limit:

65536kB

题目描述

Given three strings, you are to determine whether the third string can be formed by combining the characters in the first two strings. The first two strings can be mixed arbitrarily, but each must stay in its original order.

For example, consider forming "tcraete" from "cat" and "tree":

String A: cat
String B: tree
String C: tcraete

As you can see, we can form the third string by alternating characters from the two strings. As a second example, consider forming "catrtee" from "cat" and "tree":

String A: cat
String B: tree
String C: catrtee

Finally, notice that it is impossible to form "cttaree" from "cat" and "tree".

输入

The first line of input contains a single positive integer from 1 through 1000. It represents the number of data sets to follow. The processing for each data set is identical. The data sets appear on the following lines, one data set per line.

For each data set, the line of input consists of three strings, separated by a single space. All strings are composed of upper and lower case letters only. The length of the third string is always the sum of the lengths of the first two strings. The first two strings will have lengths between 1 and 200 characters, inclusive.

输出

For each data set, print:

Data set n: yes

if the third string can be formed from the first two, or

Data set n: no

if it cannot. Of course n should be replaced by the data set number. See the sample output below for an example.

样例输入

3
cat tree tcraete
cat tree catrtee
cat tree cttaree

样例输出

Data set 1: yes
Data set 2: yes
Data set 3: no

Global No.

1194

#include<stdio.h>
#include<string.h>
bool a[201][201];
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        memset(a,false,sizeof(a));
        char s1[201],s2[201],s[405];
        scanf("%s%s%s",s1,s2,s);
        int l1=strlen(s1);
        int l2=strlen(s2);
        for(int i=1;i<=l1;i++)
        if(s1[i-1]==s[i-1]) a[i][0]=true;
        for(int j=1;j<=l2;j++)
        if(s2[j-1]==s[j-1]) a[0][j]=true;
        for(int i=1;i<=l1;i++)
          for(int j=1;j<=l2;j++)
          if(!a[i-1][j]&&!a[i][j-1]) a[i][j]=false;
          else
          if(a[i-1][j]&&s[i+j-1]==s1[i-1]) a[i][j]=true;
          else
          if(a[i][j-1]&&s[i+j-1]==s2[j-1]) a[i][j]=true;
          else
          a[i][j]=false;
        if(a[l1][l2]) printf("Data set %d: yes/n",cas++);
        else printf("Data set %d: no/n",cas++);
    }
    return 0;
}