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  • HDU-5510 Bazinga

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 2390    Accepted Submission(s): 746

    Problem Description
    Ladies and gentlemen, please sit up straight.
    Don't tilt your head. I'm serious.

    For n given strings S1,S2,,Sn, labelled from 1 to n, you should find the largest i (1in) such that there exists an integer j (1j<i) and Sj is not a substring of Si.

    A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
     
    Input
    The first line contains an integer t (1t50) which is the number of test cases.
    For each test case, the first line is the positive integer n (1n500) and in the following n lines list are the strings S1,S2,,Sn.
    All strings are given in lower-case letters and strings are no longer than 2000 letters.
     
    Output
    For each test case, output the largest label you get. If it does not exist, output 1.
     
    Sample Input
    4
    5
    ab
    abc
    zabc
    abcd
    zabcd
    4
    you
    lovinyou
    aboutlovinyou
    allaboutlovinyou
    5
    de
    def
    abcd
    abcde
    abcdef
    3
    a
    ba
    ccc
     
    Sample Output
    Case #1: 4
    Case #2: -1
    Case #3: 4
    Case #4: 3

    题意:

    求i最大的且其前无子串的字符串,若无则输出-1。

    我们运用了求子串函数strstr();   关于strstr();的介绍请见:http://www.cnblogs.com/Kiven5197/p/5869909.html

    先找两个相邻 若找出不匹配的串再反向对比。

    附AC代码:

     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 
     4 //strstr();
     5 
     6 char s[510][2010];
     7 
     8 int main(){
     9     int t,n;
    10     scanf("%d",&t);//cin>>t;
    11     int ans=1;
    12     while(t--){
    13         scanf("%d",&n);//cin>>n;
    14         for(int i=1;i<=n;i++){
    15             scanf("%s",s[i]);//cin>>s[i];
    16         }
    17         int res=-1;
    18         for(int i=n;i>0;i--){
    19             if(!strstr(s[i],s[i-1])){
    20                 res=max(i,res);
    21                 for(int j=i+1;j<=n;j++){
    22                     if(!strstr(s[j],s[i-1])){
    23                         res=max(j,res);
    24                     }
    25                 }
    26             }
    27         }
    28         printf("Case #%d: %d
    ",ans++,res);//cout<<"Case #"<<ans++<<": "<<res<<endl;
    29     }
    30     return 0;
    31 }

    这题第一发用的cin和coutT掉了..改成scanf printf就A了

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  • 原文地址:https://www.cnblogs.com/Kiven5197/p/5869900.html
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