HDU 3518 Boring counting 后缀自动机

题目大意:

就是对于给出的字符串S, 长度不超过1000, 求其中本质不同的子串的数量, 这些子串满足在字符串S中出现了至少不重合的2次


大致思路:

如果将S建立后缀自动机的话, 记录每个节点处的串最早一次和最后一次出现的位置, 然后状态 t 处的字符串长度为[Min(t), Max(t)]比较其和rightmost - leftmost 的大小即可知道这个状态下有多少满足条件的串

时间复杂度O(n)


代码如下:

Result  :  Accepted     Memory  :  1904 KB     Time  :  15 ms

/*
 * Author: Gatevin
 * Created Time:  2015/5/4 23:07:10
 * File Name: Rin_Tohsaka.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;
#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)
#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl

#define maxm 1010
#define maxn 2020

struct Suffix_Automation
{
    struct State
    {
        State *par;
        State *go[26];
        int right, val, mi, leftmost, rightmost;
        void init(int _val = 0)
        {
            par = 0, val = _val, right = 0, mi = 0, leftmost = 1e9, rightmost = -1;
            memset(go, 0, sizeof(go));
        }
    };
    State *last, *cur, *root;
    State nodePool[maxn];
    State* newState(int val = 0)
    {
        cur->init(val);
        return cur++;
    }
    void init()
    {
        cur = nodePool;
        root = newState();
        last = root;
    }
    void extend(int w, int pos)
    {
        State *p = last;
        State *np = newState(p->val + 1);
        np->right = 1;
        np->rightmost = np->leftmost = pos;
        while(p && p->go[w] == 0)
        {
            p->go[w] = np;
            p = p->par;
        }
        if(p == 0)
        {
            np->par = root;
        }
        else
        {
            State *q = p->go[w];
            if(p->val + 1 == q->val)
            {
                np->par = q;
            }
            else
            {
                State *nq = newState(p->val + 1);
                memcpy(nq->go, q->go, sizeof(q->go));
                nq->par = q->par;
                q->par = nq;
                np->par = nq;
                while(p && p->go[w] == q)
                {
                    p->go[w] = nq;
                    p = p->par;
                }
            }
        }
        last = np;
    }
    int d[maxm];
    State* b[maxn];
    void topo()
    {
        int maxVal = 0;
        int cnt = cur - nodePool;
        memset(d, 0, sizeof(d));
        for(int i = 1; i < cnt; i++)
            maxVal = max(maxVal, nodePool[i].val), d[nodePool[i].val]++;
        for(int i = 1; i <= maxVal; i++) d[i] += d[i - 1];
        for(int i = 1; i < cnt; i++) b[d[nodePool[i].val]--] = &nodePool[i];
        b[0] = root;
    }
    void SAMInfo()
    {
        int cnt = cur - nodePool;
        State *p;
        for(int i = cnt - 1; i > 0; i--)
        {
            p = b[i];
            p->par->right += p->right;
            p->mi = p->par->val + 1;
            p->par->leftmost = min(p->par->leftmost, p->leftmost);
            p->par->rightmost = max(p->par->rightmost, p->rightmost);
        }
        return;
    }
};

Suffix_Automation sam;

char s[maxm];

int main()
{
    while(scanf("%s", s) != EOF)
    {
        if(s[0] == '#') break;
        int len = strlen(s);
        sam.init();
        for(int i = 0; i < len; i++)
            sam.extend(s[i] - 'a', i);
        sam.topo();
        sam.SAMInfo();
        int ans = 0;
        int cnt = sam.cur - sam.nodePool;
        for(int i = 1; i < cnt; i++)
            if(sam.b[i]->right >= 2)
            {
                int tmp = sam.b[i]->rightmost - sam.b[i]->leftmost;
                if(tmp >= sam.b[i]->val)
                    ans += sam.b[i]->val - sam.b[i]->mi + 1;
                else if(tmp >= sam.b[i]->mi)
                    ans += tmp - sam.b[i]->mi + 1;
            }
        printf("%d\n", ans);
    }
    return 0;
}


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