poj 2774 Long Long Message 后缀数组LCP理解

题目链接

题意:给两个长度不超过1e5的字符串，问两个字符串的连续公共子串最大长度为多少?

思路:两个字符串连接之后直接后缀数组+LCP，在height中找出max同时满足一左一右即可；

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<time.h>
using namespace std;
typedef long long ll;
const int MAXN = ;
char s[MAXN],str[MAXN];
int sa[MAXN],t[MAXN],t2[MAXN],c[MAXN],n;
void build_sa(int m,int n) // m为字符ASCII码的最大值+1;n = strlen(s) + 1;
{
    int i,*x = t, *y = t2;
    for(i = ;i < m; i++) c[i] = ;
    for(i = ;i < n; i++) c[x[i] = s[i]]++;
    for(i = ;i < m; i++) c[i] += c[i-];
    for(i = n - ;i >= ; i--) sa[--c[x[i]]] = i;
    for(int k = ;k <= n;k <<= ){
        int p = ;
        for(i = n - k;i < n;i++) y[p++] = i;
        for(i = ;i < n;i++) if(sa[i] >= k) y[p++] = sa[i] - k;

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

        swap(x,y);
        x[sa[]] = ;// 将字符彻底转变为序号;
        for(i = ,p = ;i < n;i++)
            x[sa[i]] = y[sa[i]] == y[sa[i-]] && y[sa[i]+k] == y[sa[i-]+k]?p-:p++;
        if(p >= n) break;
        m = p;
    }
}
int rk[MAXN],height[MAXN];
void getHeight()
{
    int i,j,k = ;
    for(i = ;i <= n;i++) rk[sa[i]] = i; // rk[i]:后缀i在sa[]中的下标
    for(i = ;i < n;i++){
        if(k) k--;
        if(rk[i] == ) continue;
        j = sa[rk[i] - ];
        while(i+k<n && j+k<n && s[i+k] == s[j+k]) k++;
        height[rk[i]] = k; // h[i] = height[rk[i]]; h[i] >= h[i-1] - 1;
    }
}
int main()
{
    while(scanf("%s%s",s,str) == ){
        int len = strlen(s);
        s[len] = '#';s[len+] = '\0';
        strcat(s,str);
        n = strlen(s);
        s[n] = '#';
        build_sa('z'+,n+);
        getHeight();
        int ans = -;
        for(int i = ;i <= n;i++)
            if(height[i] > ans && ((sa[i] < len && sa[i-] > len) || (sa[i] > len && sa[i-] < len)))
                ans = height[i];
        printf("%d\n",ans);
    }
    return ;
}