Substrings
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5022 Accepted Submission(s): 2253
Problem Description
You 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.
Input
The 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.
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.
Output
There should be one line per test case containing the length of the largest string found.
Sample Input
2 3 ABCD BCDFF BRCD 2 rose orchid
Sample Output
2 2
Author
Asia 2002, Tehran (Iran), Preliminary
Recommend
Eddy
//找出长度最短的串来求子串,长度由大到小
#include<stdio.h> #include<string.h> char ch[101][101]; void sub(char *p1,char *p2) { int d=strlen(p1); int i; for(i=0;i<d;i++) { p2[i]=p1[d-i-1]; } p2[i]=0; } int main(void) { int t; char c1[101]; char c2[101]; scanf("%d",&t); while(t--) { int n,i,j,k,min=101,tmp,ok=0,d; scanf("%d",&n); for(i=1;i<=n;i++) { scanf("%s",ch[i]); d=strlen(ch[i]); if(d<min) { tmp=i; min=d; } } for(i=min;i>=1 && !ok;i--) { for(j=0;j+i<=min && !ok;j++) { memcpy(c1,ch[tmp]+j,i); c1[j+i]=0; sub(c1,c2); for(k=1;k<=n;k++) { if(strstr(ch[k],c1)==NULL && strstr(ch[k],c2)==NULL) break; } if(k==n+1) { ok=1; printf("%d\n",i); } } } if(!ok) printf("0\n"); } return 0; }