POJ2774 Long Long Message

题目链接：http://poj.org/problem?id=2774

The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him these days: his mother is getting ill. Being worried about spending so much on railway tickets (Byterland is such a big country, and he has to spend 16 shours on train to his hometown), he decided only to send SMS with his mother...

题目大意：给两个字符串，求出该两个字符串中公共子串的最大长度。

思路：

1.其实我认为可以枚举第一个字符串的各个子串，然后用kmp求得公共子串的最大长度的。但好像没人这么写，是哪儿有问题吗? 如果有有缘人看到了并且知道为什么 ，欢迎加我QQ:709401953 ，帮我解答 请你吃饭饭。

2.用一个不会出现的字符连接两个字符串，对连接后的字符串跑一遍height数组，height数组的含义是字典序排名为 i 与排名为 i - 1的后缀的最长公共前缀的长度。

跑height数组模板：

void get_H()
{
    int k = 0;
    for(int i = 1; i <= len; i ++)
        rank[sa[i]] = i;
    for(int i = 1; i <= len; i ++)
    {
        if(k)
            k --;
        int j = sa[rank[i] - 1];
        while(s[i + k] == s[j + k])
            k ++;
        height[rank[i]] = k;
    }
}

AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define mem(a, b) memset(a, b, sizeof(a))
const int MAXN = 1e5 + 10;
using namespace std;

char s1[MAXN], s2[MAXN], s[2 * MAXN];
int len, size;
int rank[2 * MAXN], sa[2 * MAXN], tax[2 * MAXN], tp[2 * MAXN], height[2 * MAXN], p;

void Q_sort()
{
    for(int i = 0; i <= size; i ++)    tax[i] = 0;
    for(int i = 1; i <= len; i ++)    tax[rank[i]] ++;
    for(int i = 1; i <= size; i ++)    tax[i] += tax[i - 1];
    for(int i = len; i >= 1; i --)    sa[tax[rank[tp[i]]] --] = tp[i];
}

void get_H()
{
    int k = 0;
    for(int i = 1; i <= len; i ++)
        rank[sa[i]] = i;
    for(int i = 1; i <= len; i ++)
    {
        if(k)
            k --;
        int j = sa[rank[i] - 1];
        while(s[i + k] == s[j + k])
            k ++;
        height[rank[i]] = k;
    }
}

void get_SA()
{
    for(int i = 1; i <= len; i ++)    rank[i] = s[i], tp[i] = i;
    Q_sort();
    for(int w = 1; w <= len; w *= 2)
    {
        p = 0;
        for(int i = 1; i <= w; i ++)    tp[++ p] = len - w + i;
        for(int i = 1; i <= len; i ++)    if(sa[i] > w) tp[++ p] = sa[i] - w;
        Q_sort();
        swap(tp, rank);//tp数组无用 继承上一轮rank排名
        rank[sa[1]] = p = 1;
        for(int i = 2; i <= len; i ++)
            rank[sa[i]] = (tp[sa[i - 1]] == tp[sa[i]] && tp[sa[i - 1] + w] == tp[sa[i] + w]) ? p : ++ p;
        size = p;
        if(p >= len)
            break;
    }
    get_H();
}

int main()
{
    scanf("%s%s", s1 + 1, s2 + 1);
    int len1 = strlen(s1 + 1), len2 = strlen(s2 + 1);
    int cnt = 0;
    for(int i = 1; i <= len1; i ++)
        s[++ cnt] = s1[i];
    s[++ cnt] = '#';  //用一个不会出现的字符连接两个字符串。 
    for(int i = 1; i <= len2; i ++)
        s[++ cnt] = s2[i];
    len = strlen(s + 1), size = 130;
    get_SA();
    int ans = 0;
    for(int i = 2; i <=len; i ++)
    {
        if(sa[i - 1] <= len1 && sa[i] > len1 + 1)
            ans = max(ans, height[i]);
        else if(sa[i - 1] > len1 + 1 && sa[i] <= len1)
            ans = max(ans, height[i]);
    }
    printf("%d
", ans);
    return 0;
}