HDU1238——Substrings【KMP,列舉】
Substrings
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12486 Accepted Submission(s): 6001
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.
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
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題目大意:給你一組字串,問這組字串的最長公共子串是多長,並且如果這個子串的反轉是字串的子串則也可認為是他的子串。
大致思路:我們可以列舉第一個字串的所有子串,然後用KMP演算法看在其他字串中能否找到他或者他的反轉。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 100010;
string s[MAXN];
int NextVal[MAXN];
void GetNextVal(string a){
NextVal[0] = -1;
for(int i = 1,j = -1; i < a.size(); i++){
while(j > -1 && a[i] != a[j + 1]) j = NextVal[j];
if(a[i] == a[j + 1]) j++;
NextVal[i] = j;
}
}
bool KMP(string a,string b){
for(int i = 0, j = -1; i < b.size(); i++){
while(j > -1 && (j == a.size() - 1 || b[i] != a[j + 1])) j = NextVal[j];
if(b[i] == a[j + 1]) j++;
if(j == a.size() - 1){
return true;
}
}
return false;
}
int main(int argc, char const *argv[])
{
int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
for(int i = 0; i < n; i++)
cin >> s[i];
string ans = "";
for(int i = 0; i < s[0].size();i++){
for(int j = 1; j <= s[0].size() - i; j++){
string op1 = s[0].substr(i,j);
string op2 = op1;
reverse(op2.begin(),op2.end());
bool flag1 = true,flag2 = true;
GetNextVal(op1);
for(int k = 1; k < n; k++){
if(!KMP(op1,s[k])){ flag1 = false; break; }
}
GetNextVal(op2);
for(int k = 1; k < n; k++){
if(!KMP(op2,s[k])){ flag2 = false; break;}
}
if(flag1 || flag2){
if(ans.size() <= op1.size()) ans = op1;
}
}
}
//cout << ans << endl;
cout << ans.size() << endl;
}
return 0;
}