poj 2774

http://poj.org/problem?id=2774

题意：求两个字符串的最大公共子串

思路:SAM模板

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
#define maxn 100010

char str[maxn];

struct Node{
    int len, link, nex[26];
}st[maxn << 1];
int root, size, last;

void init(){
    root = size = last = 0;
    st[root].len = 0;
    st[root].link = -1;
}

void Extend(int c){
    int p = last, cur = ++ size;
    st[cur].len = st[p].len + 1;
    for(; ~p && st[p].nex[c] == 0; p = st[p].link)
        st[p].nex[c] = cur;
    if(p == -1)
        st[cur].link = root;
    else{
        int q = st[p].nex[c];
        if(st[q].len == st[p].len + 1)
            st[cur].link = q;
        else{
            int clone = ++ size;
            st[clone] = st[q];
            st[clone].len = st[p].len + 1;
            for(; ~p && st[p].nex[c] == q; p = st[p].link)
                st[p].nex[c] = clone;
            st[q].link = st[cur].link = clone;
        }
    }
    last = cur;
}




int main()
{
    init();
    scanf("%s",str);
    int len =strlen(str);
    for(int i = 0;i<len;i++)
        Extend(str[i]-'a');
    scanf("%s",str);
    len =strlen(str);
    int n = root,cur = 0,ans = 0;
    for(int i = 0;i<len;i++)
    {
        int tmp = str[i]-'a';
        if(st[n].nex[tmp])
            cur++,n = st[n].nex[tmp];
        else {
            while(~n&&st[n].nex[tmp]==0)   //字符串段了，从前一个进行匹配
                n = st[n].link;
            if(n==-1)
                n = root,cur = 0;     //是否为第一个字符
            else cur = st[n].len+1,n = st[n].nex[tmp];
        }
        ans = max(ans,cur);
    }
    printf("%d
",ans);
    return 0;
}