hdu4763

maksyuki 发表于 oj 分类，标签: 字符串-KMP

15 9月 2015

Theme Section

Problem Description

It's time for music! A lot of popular musicians are invited to join us in the music festival. Each of them will play one of their representative songs. To make the programs more interesting and challenging, the hosts are going to add some constraints to the rhythm of the songs, i.e., each song is required to have a 'theme section'. The theme section shall be played at the beginning, the middle, and the end of each song. More specifically, given a theme section E, the song will be in the format of 'EAEBE', where section A and section B could have arbitrary number of notes. Note that there are 26 types of notes, denoted by lower case letters 'a' - 'z'.

To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?

Input

The integer N in the first line denotes the total number of songs in the festival. Each of the following N lines consists of one string, indicating the notes of the i-th (1 <= i <= N) song. The length of the string will not exceed 10^6.

Output

There will be N lines in the output, where the i-th line denotes the maximum possible length of the theme section of the i-th song.

Sample Input

abc

aaa

aaaaba

aaxoaaaaa

Sample Output

Source

2013 ACM/ICPC Asia Regional Changchun Online

题目类型：KMP

算法分析：使用next数组从大到小枚举相等的前后缀的长度，然后在中间再使用一次KMP构建一个next，然后进行一次匹配，直到找到结果或者返回0

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

const int INF = 0x7FFFFFFF;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1000000 + 66;

char s[maxn], pp[maxn];
int Next[maxn];

void Get (char *s)
{
    int lens = strlen (s);
    Next[0] = -1, Next[1] = 0;
    int k = 0;
    for (int i = 2; i <= lens; i++)
	{
        while (k != -1 && s[k] != s[i-1])
			k = Next[k];

        Next[i] = ++k;
	}
}

bool Solve (char *s, char *p, int sa, int sb)
{
    int lenp = strlen (p);
    int i = sa, j = 0;
    while (i <= sb)
	{
        if (j == -1 || s[i] == p[j])
			i++, j++;

		else
			j = Next[j];

		if (j == lenp)
			return true;
	}

return false;
}

int main()
{
//	ifstream cin("aaa.txt");
//	freopen ("aaa.txt", "r", stdin);

    int t;
    cin >> t;

    while (t--)
	{
        cin >> s;
		Get (s);

		int len = strlen (s);
        int po = Next[len];

		bool is_valid = false;
        while (po)
		{
            for (int i = 0; i < po; i++)
				pp[i] = s[i];
			pp[po] = 0;

			Get (pp);
            if (Solve (s, pp, po, len - po - 1))
				break;

			po = Next[po];
		}

		cout << po << endl;
	}
return 0;
}

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109

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

const int INF = 0x7FFFFFFF;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1000000 + 66;

char s[maxn], pp[maxn];

int Next[maxn];

void Get (char *s)

{

int lens = strlen (s);

Next[0] = -1, Next[1] = 0;

int k = 0;

for (int i = 2; i <= lens; i++)

{

while (k != -1 && s[k] != s[i-1])

k = Next[k];

Next[i] = ++k;

}

bool Solve (char *s, char *p, int sa, int sb)

{

int lenp = strlen (p);

int i = sa, j = 0;

while (i <= sb)

{

if (j == -1 || s[i] == p[j])

i++, j++;

else

j = Next[j];

if (j == lenp)

return true;

}

return false;

}

int main()

{

// ifstream cin("aaa.txt");

// freopen ("aaa.txt", "r", stdin);

int t;

cin >> t;

while (t--)

{

cin >> s;

Get (s);

int len = strlen (s);

int po = Next[len];

bool is_valid = false;

while (po)

{

for (int i = 0; i < po; i++)

pp[i] = s[i];

pp[po] = 0;

Get (pp);

if (Solve (s, pp, po, len - po - 1))

break;

po = Next[po];

}

cout << po << endl;

}

return 0;

}

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