【题解】HDU-4763 Theme Section

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HDU-4763 Theme Section 类型 KMP next数组性质

Theme Section(HDU-4763)

题面

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?

输入

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.

输出

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.

样例输入

1
2
3
4
5
6
5
xy
abc
aaa
aaaaba
aaxoaaaaa

样例输出

1
2
3
4
5
0
0
1
1
2

提示

无

思路

标记第一个E的结尾，枚举第二个E，满足条件的求个max即可。数据有点水。

代码

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47
char t[mxn];
int nxt[mxn], vis[mxn];

void getnxt(char* t, int m)
{
    int i = 0, j = -1; nxt[0] = -1;
    while (i < m)
    {
        if (j == -1 || t[i] == t[j]) {
            i++, j++;
            // if (t[i] == t[j])
            //     nxt[i] = nxt[j]; // next数组优化
            // else
                nxt[i] = j;
        } else
            j = nxt[j];
    }
}

int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%s", t);
        int n = strlen(t);
        for(int i=0; i<n; i++) vis[i] = 0;
        getnxt(t, n);

        for(int i=n; i>0; i=nxt[i]){
            if(3*i<=n)
                vis[i] = 1;
        }

        int ans=0;
        for(int i=n-1; i>1; i--) {
            for(int j=i; j>0; j=nxt[j]) {
                if(vis[j] && 2*j<=i && i+j<=n) {
                    ans = max(ans, j);
                    break;
                }
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}