SETI--后缀数组

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Problem Description

      Amateur astronomers Tom and Bob try to find radio broadcasts of extraterrestrial civilizations in the air. Recently they received some strange signal and represented it as a word consisting of small letters of the English alphabet. Now they wish to decode the signal. But they do not know what to start with.
      They think that the extraterrestrial message consists of words, but they cannot identify them. Tom and Bob call a subword of the message a potential word if it has at least two non-overlapping occurrences in the message.

      For example, if the message is “abacabacaba”, “abac” is a potential word, but “acaba” is not because two of its occurrences overlap.
      Given a message m help Tom and Bob to find the number of potential words in it.

Input

      Input file contains one string that consists of small letters of the English alphabet. The length of the message doesn’t exceed 10 000.

Output

      Output one integer number — the number of potential words in a message.

Sample Input

abacabacaba

Sample Output

15

思路：后缀数组搞。len表示之前最长的已经贡献种数的长度。如果当前height[i]<len。那么是不能贡献新串的，这时更新len。如果height[i]>len，用min(abs(sa[i]-sa[i-1]),height[i])尝试去更新len。注意还得往前面跑，虽然往前面跑LCP肯定是减少，但是sa[i]-sa[pos]可能变大。导致min()变大。
#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <iomanip>#include <cassert>using namespace std;#pragma comment(linker, "/stack:1024000000,1024000000")#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define travel(e, u) for(int e = u, v = vv[u]; e; e = nxt[e], v = vv[e])#define bit(n) (1LL<<(n))#define And(a, b) ((a) & (b))#define Xor(a, b) ((a) ^ (b))#define Or(a, b) ((a) | (b))#define mid ((l+r)/2)#define ls (pos<<1)#define rs (ls|1)#define lson l, m, ls#define rson m+1, r, rstypedef long long LL;typedef unsigned long long ULL;void work();int main(){#ifdef ACM    freopen("in.txt", "r", stdin);#endif // ACM     work();} void scanf(int & x, char c = 0){    while((c = getchar()) < '0' || c > '9');     x = c - '0';    while((c = getchar()) >= '0' && c <= '9') x = x * 10 + (c - '0');} #define rep(i,a,b) for(int i = (a);i<(b);++i)#define rrep(i,a,b) for(int i = (a);i>=(b);--i)#define clr(a,x) memset(a,x,sizeof(a))/*****************************************华丽分割线**********************************************/#define maxn 20008int str[maxn],vis[maxn];char s[maxn];int sa[maxn],t[maxn],t2[maxn],c[maxn],key[maxn];int height[maxn],Rank[maxn];void build_sa(int * s,int n,int m){    int i,*x = t,*y = t2;    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];        //根据sa和y数组计算新的数y组        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;    }} void getHeight(int * s,int n){    int i,j,k = 0;    for(i = 0;i < n;i++)    Rank[sa[i]] = i;    for(i = 0;i < n;i++)    {        if(k) k--;        int j = sa[Rank[i]-1];        while(s[i+k] == s[j+k])        k++;        height[Rank[i]] = k;    }} LL gao(int n){    LL ans = 0;    int len = 0;    for(int i = 1;i < n;i++)    {        if(height[i] > len)        {            int pos = i-1,fuck = height[i],lcp = height[i];            while(1)            {                fuck = min(abs(sa[i]-sa[pos]),lcp);                if(lcp > len)                {                    if(fuck > len)                    {                        ans += fuck - len;                        len = fuck;                    }                    pos--;                    if(pos == 0)    break;                }                else break;                lcp = min(lcp,height[pos+1]);            }        }        else len = height[i];    }    return ans;}void work(){    while(scanf("%s",s)!=EOF)    {        int len = strlen(s);        for(int i = 0;i < len;i++)            str[i] = s[i] - 'a' + 2;        str[len] = 0;        build_sa(str,len+1,30);        getHeight(str,len+1);        LL ans = gao(len+1);        printf("%lld\n",ans);    }}