SPOJ

题目：求2个字符串的最长公共子串的长度

思路：对str1建立后缀自动机，然后让str2去跑

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fconst int MAXN = 500050, SIZE = 26;struct SAM {    int len[MAXN], link[MAXN], next[MAXN][SIZE];    int total, last;    inline int newNode(int L) {        len[++total] = L; link[total] = 0;        for(int i = 0; i < SIZE; ++i) next[total][i] = 0;        return total;    }    inline void Add(int c) {        int i, p = last, cur = newNode(len[last] + 1);        for(; p && !next[p][c]; p = link[p]) next[p][c] = cur;        if(!p) link[cur] = 1;//令其指向初始状态        else {            int q = next[p][c];            if(len[q] == len[p] + 1) link[cur] = q;            else {//>                int clone = newNode(len[p] + 1);                for(i = 0; i < SIZE; ++i) next[clone][i] = next[q][i];                link[clone] = link[q];                link[q] = link[cur] = clone;for(; p && next[p][c] == q; p = link[p]) next[p][c] = clone;            }        }        last = cur;    }    void Init () {//根节点是1        total = 0;        last = newNode(0);    }}sam;char str1[MAXN],str2[MAXN];int main(){    while(~scanf("%s%s",str1,str2)){        sam.Init();        for(int i=0;str1[i]!='\0';i++)            sam.Add(str1[i]-'a');        int ans=0,len=0,now=1;        for(int i=0;str2[i]!='\0';i++){            int c=str2[i]-'a';            if(sam.next[now][c]){                len++;                now=sam.next[now][c];            }            else{                while(sam.next[now][c]==0&&now!=0)                    now=sam.link[now];                if(now==0){                    now=1;                    len=0;                }                else{                    len=sam.len[now]+1;                    now=sam.next[now][c];                }            }            ans=max(ans,len);        }        printf("%d\n",ans);    }    return 0;}