题目链接:http://poj.org/problem?id=3080
题目大意:给你N个长度为60的字符串(N<=10),求他们的最长公共子串(长度>=3)。
题目分析:KMP字符串匹配基础题。直接枚举第1个字符串的所有子串,判断这个子串是否出现在另外N-1个串中。
实现代码如下:
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 66;
int n, m, nxt[maxn];
string s, t; // s代表母串,t代表子串
int T, N;
string ss[11];
vector<string> tt;
void cal_next() {
m = t.length();
for (int i = 0, j = -1; i < m; i ++) {
while (j != -1 && t[j+1] != t[i]) j = nxt[j];
nxt[i] = (j+1 < i && t[j+1] == t[i]) ? ++j : -1;
}
}
bool check_s_has_t() {
n = s.length(); // cal_next();
for (int i = 0, j = -1; i < n; i ++) {
while (j != -1 && t[j+1] != s[i]) j = nxt[j];
if (t[j+1] == s[i]) {
j ++;
if (j >= m-1) {
return true;
}
}
}
return false;
}
bool cmp(string s, string t) {
if (s.length() != t.length()) return s.length() > t.length();
return s < t;
}
int main() {
cin >> T;
while (T --) {
cin >> N;
for (int i = 0; i < N; i ++) cin >> ss[i];
tt.clear();
for (int i = 0; i < 60; i ++) {
for (int j = 3; i+j <= 60; j ++) {
string tmp_s = ss[0].substr(i, j);
tt.push_back(tmp_s);
// cout << "tmp: " << tmp_s << endl;
}
}
sort(tt.begin(), tt.end(), cmp);
bool findOne = false;
for (int i = 0; i < tt.size(); i ++) {
t = tt[i];
cal_next();
bool flag = true;
for (int j = 1; j < N; j ++) {
s = ss[j];
if (check_s_has_t() == false) {
flag = false;
break;
}
}
if (flag == true) {
findOne = true;
break;
}
}
if (!findOne) puts("no significant commonalities");
else cout << t << endl;
}
return 0;
}
作者:zifeiy