HDU 1238 Substrings 字符串水题，STL String 的应用

SubstringsTime Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 19   Accepted Submission(s) : 10Problem DescriptionYou are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.InputThe first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.OutputThere should be one line per test case containing the length of the largest string found.Sample Input23ABCDBCDFFBRCD2roseorchidSample Output22

来源： http://acm.hdu.edu.cn/webcontest/contest_showproblem.php?pid=1000&ojid=0&cid=10646&hide=0

详解在注释中

#include <cstdio>#include <iostream>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;#define MAXN 100*2#define MAXLen 100*2struct String{    string S;    int len;}Str[MAXN];int Cal(int n){    int Max_Size = INT_MIN;    for(int i=0;i<Str[1].len;i++)    {        for(int j=i;j<Str[1].len;j++)       //枚举最小串的每个子串        {                           string S(&Str[1].S[i],&Str[1].S[j]+1);  //将S初始化为该子串            int flag = 1;            for(int m=2;m<=n;m++)           //对其他串及其倒置串暴力枚举查找            {                int Loc1 = Str[m].S.find(S);                    reverse(Str[m].S.begin(),Str[m].S.end());                int Loc2 = Str[m].S.find(S);                if(Loc1<0&& Loc2<0) flag = 0;            }            int LENS = S.size();            if(flag) Max_Size = max(Max_Size,LENS); //更新长度        }    }    return Max_Size>=0?Max_Size:0;      //防止没有公共串}int cmp(const void *a,const void *b){    return (*(String *)a).len - (*(String *)b).len;}int main(void){ //   freopen("F:\\test.txt","r",stdin);    int T;scanf("%d",&T);    for(int t=1,n;t<=T&&scanf("%d",&n);++t)    {        for(int i=1;i<=n;++i)        {            char temp[MAXLen];            scanf("%s",temp);               //定义结构体 内存放Str和它的长度，便于排序            Str[i].S = temp;            Str[i].len = Str[i].S.size();        }        qsort(Str+1,n,sizeof Str[0],cmp);   //找出最小的边        printf("%d\n",Cal(n));    }}