HDU1403 Longest Common Substring —— 后缀数组_longest common substring hdu

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1403

Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6792 Accepted Submission(s): 2401

Problem Description

Given two strings, you have to tell the length of the Longest Common Substring of them.

For example:
str1 = banana
str2 = cianaic

So the Longest Common Substring is "ana", and the length is 3.

Input

The input contains several test cases. Each test case contains two strings, each string will have at most 100000 characters. All the characters are in lower-case.

Process to the end of file.

Output

For each test case, you have to tell the length of the Longest Common Substring of them.

Sample Input

  
   banana
cianaic

Sample Output

Author

Ignatius.L

题解：

1.将两个字符串接在一起，并用特殊分隔符分开。然后跑一下后缀数组，得出height[]数组。

2.枚举height[]数组，如果排位相邻的两个后缀字符串的开头分别在两边（原先的两个字符串），则更新答案。

http://blog.youkuaiyun.com/j_sure/article/details/41777097

代码如下：

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXN = 2e5+10;

bool cmp(int *r, int a, int b, int l)
{
    return r[a]==r[b] && r[a+l]==r[b+l];
}

int t1[MAXN], t2[MAXN], c[MAXN];
void DA(int str[], int sa[], int Rank[], int height[], int n, int m)
{
    int i, j, p, *x = t1, *y = t2;
    for(i = 0; i<m; i++) c[i] = 0;
    for(i = 0; i<n; i++) c[x[i] = str[i]]++;
    for(i = 1; i<m; i++) c[i] += c[i-1];
    for(i = n-1; i>=0; i--) sa[--c[x[i]]] = i;
    for(j = 1; j<=n; j <<= 1)
    {
        p = 0;
        for(i = n-j; i<n; i++) y[p++] = i;
        for(i = 0; i<n; i++) if(sa[i]>=j) y[p++] = sa[i]-j;

        for(i = 0; i<m; i++) c[i] = 0;
        for(i = 0; i<n; i++) c[x[y[i]]]++;
        for(i = 1; i<m; i++) c[i] += c[i-1];
        for(i = n-1; i>=0; i--) sa[--c[x[y[i]]]] = y[i];

        swap(x, y);
        p = 1; x[sa[0]] = 0;
        for(i = 1; i<n; i++)
            x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;

        if(p>=n) break;
        m = p;
    }

    int k = 0;
    for(i = 0; i<=n; i++) Rank[sa[i]] = i;
    for(i = 0; i<n; i++)
    {
        if(k) k--;
        j = sa[Rank[i]-1];
        while(str[i+k]==str[j+k]) k++;
        height[Rank[i]] = k;
    }
}

char a[MAXN], b[MAXN];
int r[MAXN], sa[MAXN], Rank[MAXN], height[MAXN];
// sa[]下标的有效值：0~n-1；  height[]下标的有效值：1~n-1
int main()
{
    while(scanf("%s%s", a, b)!=EOF)
    {
        int lena, lenb, len;
        lena = strlen(a);
        lenb = strlen(b);

        len = 0;
        for(int i = 0; i<lena; i++) r[len++] = a[i];
        r[len++] = 1;
        for(int i = 0; i<lenb; i++) r[len++] = b[i];
        r[len++] = 0;
        DA(r, sa, Rank, height, len, 128);

        int ans = 0;
        for(int i = 1; i<len; i++)
            if( (sa[i]<lena && sa[i-1]>lena) || (sa[i]>lena && sa[i-1]<lena) )
                ans = max(ans, height[i]);
        printf("%d\n", ans);
    }
    return 0;
}