POJ

原题链接

题意：

求两个字符串的最长公共子串的长度。

思路：

利用后缀数组中 height[] 的意义，将两个字符串拼接为一个字符串，中间插入特殊符号；然后求得来自两个不同串的后缀的最长公共前缀即可。

/*
* @Author: WindyStreet
* @Date:   2018-07-31 14:30:43
* @Last Modified by:   WindyStreet
* @Last Modified time: 2018-07-31 14:37:47
*/
#include<bits/stdc++.h>

using namespace std;

#define X first
#define Y second
#define eps  1e-2
#define gcd __gcd
#define pb push_back
#define PI acos(-1.0)
#define lowbit(x) (x)&(-x)
#define bug printf("!!!!!
");
#define mem(x,y) memset(x,y,sizeof(x))

typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef unsigned long long uLL;

const int maxn = 2e5+7;
const int INF  = 1<<30;
const int mod  = 1e9+7;
char s[maxn],tmp[maxn];
int sa[maxn],h[maxn],rk[maxn],c[maxn],t[maxn],t2[maxn];
void work(int n,int m){
    int i, *x = t, *y = t2;
    //基数排序
    for(i = 0; i < m; i++) c[i] = 0;
    for(i = 0; i < n; i++) c[x[i]=s[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(int k = 1; k <= n; k <<= 1){
        int p = 0 ;
        for(i = n - k; i < n; i++) y[p++] = i;
        for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;
        for(i = 0; i < m; i++) c[i] = 0;
        for(i = 0; i < n; i++) c[x[y[i]]]++;
        for(i = 0; 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]] = y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k] ? p-1:p++;
         }
         if(p>=n)break;
         m = p;
    }
    int k = 0;
    for(i = 0; i < n; i++) rk[sa[i]] = i;
    for(i = 0; i < n; i++){
        if(k) k--;
        int j = sa[rk[i]-1];
        while(s[i+k]==s[j+k])k++;
        h[rk[i]] = k;
    }
}

void solve(){
    scanf("%s",s);
    scanf("%s",tmp);
    int len = strlen(s);
    int len1= strlen(tmp);
    s[len] = '#';                                 //中间插入特殊字符
    for(int i = 0;i<len1;i++){
        s[len+i+1] = tmp[i];
    }
    s[len + len1+1] = '';
    work(len+len1+1,128);
    int ans = 0;
    for(int i=2;i<=len+len1+1;i++){
        if(1LL*(sa[i] - len)*1ll*(sa[i-1] - len)<0)        //判断是否来自两个不同的串
            ans = max(ans, h[i]);
    }
    printf("%d
",ans);
    
    return;
}

int main()
{
//    freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
//    ios::sync_with_stdio(false);
    int t = 1;
    //scanf("%d",&t);
    while(t--){
    //    printf("Case %d: ",cas++);
        solve();
    }
    return 0;
}