POJ3080 && POJ 3450

题意: 给定m个串,让你找出它们的最长公共子串

思路: 先二分串的长度,枚举该长度的串(可以从第一个串里找),看该长度是否合法,(就是用这个长度的所有串去匹配给定的 第 2 - m 个串 为提高效率,用kmp)然后得到一个最大长度,再在该长度下寻找一个字典序最小的解即可

POJ3450代码(3080类似)

/*************************************************************************
    > File Name: poj3450.cpp
    > Author: ALex
    > Mail: [email protected]
    > Created Time: 2015年02月02日 星期一 20时14分57秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

const int N = 222;
int next[N];
char str[N];
char mat[4010][N];
char p[N];
int m;

void get_next ()
{
	int len = strlen (str);
	next[0] = -1;
	int j = 0, k = -1;
	while (j < len)
	{
		if (k == -1 || str[k] == str[j])
		{
			++k;
			++j;
			next[j] = k;
		}
		else
		{
			k = next[k];
		}
	}
}

bool match_ok (int s)
{
	int i = 0, j = 0;
	int lens = strlen (mat[s]);
	int lent = strlen (str);
	while (i < lens && j < lent)
	{
		if (j == -1 || str[j] == mat[s][i])
		{
			++i;
			++j;
		}
		else
		{
			j = next[j];
		}
	}
	if (j == lent)
	{
		return 1;
	}
	return 0;
}

bool solve (int L)
{
	bool flag = false;
	int len = strlen (mat[1]);
	for (int i = 0; i + L - 1 < len; ++i)
	{
		for (int j = i; j < i + L; ++j)
		{
			str[j - i] = mat[1][j];
		}
		str[L] = '\0';
		get_next ();
		bool tmp = true;
		for (int j = 2; j <= m; ++j)
		{
			if (!match_ok (j))
			{
				tmp = false;
				break;
			}
		}
		if (tmp)
		{
			flag = true;
			break;
		}
	}
	return flag;
}

void Binsearch ()
{
	int len = strlen (mat[1]);
	int l = 1, r = len, mid, ans = -1;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (solve (mid))
		{
			ans = mid;
			l = mid + 1;
		}
		else
		{
			r = mid - 1;
		}
	}
	if (ans == -1)
	{
		printf("IDENTITY LOST\n");
		return;
	}
	bool flag = false;
	for (int i = 0; i + ans <= len; ++i)
	{
		for (int j = i; j < i + ans; ++j)
		{
			str[j - i] = mat[1][j];
		}
		str[ans] = '\0';
		get_next ();
		bool x = true;
		for (int j = 2; j <= m; ++j)
		{
			if (!match_ok (j))
			{
				x = false;
				break;
			}
		}
		if (x)
		{
			if (!flag)
			{
				flag = true;
				strcpy (p, str);
			}
			else
			{
				if (strcmp (p, str) > 0)
				{
					strcpy (p, str);
				}
			}
		}
	}
	printf("%s\n", p);
}

int main ()
{
	while (~scanf("%d", &m), m)
	{
		for (int i = 1; i <= m; ++i)
		{
			scanf("%s", mat[i]);
		}
		Binsearch ();
	}
	return 0;
}