zoukankan      html  css  js  c++  java
  • HDU-1501 Zipper

    http://acm.hdu.edu.cn/showproblem.php?pid=1501

    Zipper

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5810    Accepted Submission(s): 2088

    Problem Description
    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".
     
    Input
    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.
     
    Output
    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.
     
    Sample Input
    3
    cat tree tcraete
    cat tree catrtee
    cat tree cttaree
     
    Sample Output
    Data set 1: yes
    Data set 2: yes
    Data set 3: no
    #include<stdio.h>  
    #include<string.h>  
    char str1[205],str2[205],str3[405];  
    int flag;  
    int len1,len2,len3;  
    int mark[205][205];  
    void dfs(int a,int b,int c)  
    {  
        if(flag)  
        return;  
        if(c==len3)  
        {  
            flag=1;  
            return;  
        }  
         if(mark[a][b]==1)  
            return;  
          mark[a][b]=1;  
        if(str1[a]==str3[c])  
        dfs(a+1,b,c+1);  
        if(str2[b]==str3[c])  
        dfs(a,b+1,c+1);  
    }  
    int main()  
    {  
        int i,t;  
        scanf("%d",&t);  
        for(i=1;i<=t;i++)  
        {  
            scanf("%s%s%s",str1,str2,str3);  
            printf("Data set %d: ",i);  
            len1=strlen(str1);  
            len2=strlen(str2);  
            len3=strlen(str3);  
            if(len1+len2!=len3)  
            {  
                printf("no
    ");  
                continue;  
            }  
            memset(mark,0,sizeof(mark));  
            flag=0;  
            dfs(0,0,0);  
            if(flag)  
            printf("yes
    ");  
            else  
            printf("no
    ");  
        }  
        return 0;  
    }  
    
  • 相关阅读:
    Mysql update from
    抽象类
    表自链接递归查询死循环
    复制订阅服务器和 AlwaysOn 可用性组 (SQL Server)
    C#找出接口的所有实现类并遍历执行这些类的公共方法
    Cors Http 访问控制
    返回参数去掉xml格式,以纯json格式返回(转)
    混布技术提升资源利用率
    fair scheduler配置
    ambari安装
  • 原文地址:https://www.cnblogs.com/cancangood/p/3477484.html
Copyright © 2011-2022 走看看