PKU 3450 - Corporate Identity (二分 + 暴力)
题意
找出各个字符串中共同的子串,最长。
思路
枚举第一个字符串的子串,然后在之后的字符串中找。
枚举子串的时候可以用二分减少复杂度。当找到某个长度的时候再在那个长度里找字典序最小的。
代码
#include <cstdio>
#include <stack>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
#define LL long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define F first
#define S second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1
#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
using namespace std;
const int MAXN = 4000 + 10;
const int MOD = 1000007;
typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
string str[MAXN], ans;
int n;
bool flag;
bool Check(int mid)
{
string sub;
int j = 0;
for (int i = 0; i + mid <= SZ(str[0]); i++)
{
sub = str[0].substr(i, mid);
for (j = 1; j < n; j++)
if (str[j].find(sub) == string::npos) break;
if (j == n)
{
flag = true;
ans = sub;
return true;
}
}
return false;
}
void Output()
{
string sub;
int j;
for (int i = 0; i + SZ(ans) <= SZ(str[0]); i++)
{
sub = str[0].substr(i, SZ(ans));
for (j = 1; j < n; j++)
if (str[j].find(sub) == string::npos) break;
if (j == n) ans = min(ans, sub);
}
cout << ans << endl;
}
void Solve()
{
int l = 1, r = 200, mid;
flag = false;
while (l <= r)
{
mid = MID(l, r);
if (mid == 0) break;
if (Check(mid)) l = mid + 1;
else r = mid - 1;
}
if (flag) Output();
else cout << "IDENTITY LOST" << endl;
}
int main()
{
//ROP;
ios::sync_with_stdio(0);
int i, j;
while (cin >> n, n)
{
for (i = 0; i < n; i++) cin >> str[i];
Solve();
}
return 0;
}