POJ 3415

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Common Substrings

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 8358 Accepted: 2761

Description

A substring of a string T is defined as:

T(i, k)=T_iT_i₊₁...T_i+k_-1, 1≤i≤i+k-1≤|T|.

Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):

S = {(i, j, k) | k≥K, A(i, k)=B(j, k)}.

You are to give the value of |S| for specific A, B and K.

Input

The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.

1 ≤ |A|, |B| ≤ 10⁵
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.

Output

For each case, output an integer |S|.

Sample Input

2aababaaabaabaa1xxxx0

Sample Output

Source

POJ Monthly--2007.10.06, wintokk

题意：给出两个字符串SA, SB和一个数K，求两个字符串的长度不小于k的公共子串个数。

思路：两个字符串中间用一个没出现过的字符连接，之后求出height数组，根据K将后缀分组，对每一组判断每个后缀属于SA串还是SB串，对每个SA串，求前面的SB串与它的LCP长度，同样，对每个SB串，求前面的SA串与它的LCP长度。对于两个后缀，设它们的LCP值为L，则这两个后缀能产生L-K+1个公共字串。而求每个SA串与前面的SB串能产生多少公共子串可以用单调栈O(len)的实现：栈中维护每个后缀的height值，以及从起点扫描到该位置时，有多少个后缀的height值大于等于当前后缀的height值，于是每次扫描到一个后缀，将其加入单调栈中(需要弹出小于等于它的元素)，计算该后缀的cnt值。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <string>#include <algorithm>#include <map>#include <stack>using namespace std;#define INF 110000#define N 201000#define M 201000int k, n, m, len1, len2;char s[N];int str[N], rank[N], sa[N], bucket[N], trank[N], sa2[N], height[N];class Node{public:    int val, cnt;    Node(int _val, int _cnt)    {        val = _val;        cnt = _cnt;    }};stack<Node> sta;int cmp(int p1, int p2, int len){    return p1+len < n && p2+len < n && trank[p1] == trank[p2] && trank[p1+len] == trank[p2+len];}void getSa(){    for(int i = 0; i < m; i++) bucket[i] = 0;    for(int i = 0; i < n; i++) bucket[rank[i] = str[i]]++;    for(int i = 1; i < m; i++) bucket[i] += bucket[i-1];    for(int i = n-1; i >= 0; i--) sa[--bucket[rank[i]]] = i;    for(int j = 1, p = 0; p < n; j <<= 1, m = p)    {        p = 0;        for(int i = n-j; i < n; i++) sa2[p++] = i;        for(int i = 0; i < n; i++)            if(sa[i] >= j) sa2[p++] = sa[i]-j;        for(int i = 0; i < m; i++) bucket[i] = 0;        for(int i = 0; i < n; i++) bucket[rank[i]]++;        for(int i = 1; i < m; i++) bucket[i] += bucket[i-1];        for(int i = n-1; i >= 0; i--) sa[--bucket[rank[sa2[i]]]] = sa2[i];        for(int i = 0; i < n; i++) trank[i] = rank[i];        p = 0; rank[sa[0]] = p++;        for(int i = 1; i < n; i++)            rank[sa[i]] = cmp(sa[i], sa[i-1], j) ? p-1 : p++;    }}void getHeight(){    int h = 0;    height[0] = 0;    for(int i = 0; i < n; i++)    {        if(rank[i] == 0) continue;        if(h) h--;        int pre = sa[rank[i]-1];        for(; pre+h < n && i+h < n && str[i+h] == str[pre+h]; h++);        height[rank[i]] = h;    }}inline int get(int pos){    if(sa[pos] == len1) return 0;    return sa[pos] < len1 ? 1 : 2;}long long cal(int l, int r, int flag){    while(!sta.empty()) sta.pop();    long long res = 0, sum = 0;    if(get(l) != flag)    {        sta.push(Node(INF, 1));        sum += INF-k+1;    }    for(int i = l+1; i <= r; i++)    {        int cnt = 0;        while(!sta.empty() && sta.top().val >= height[i])        {            sum -= (sta.top().val-k+1)*sta.top().cnt;            cnt += sta.top().cnt;            sum += (height[i]-k+1)*sta.top().cnt;            sta.pop();        }        if(cnt)            sta.push(Node(height[i], cnt));        if(get(i) != flag)        {            sum += INF-k+1;            sta.push(Node(INF, 1));        }        if(get(i) == flag)        {            res += sum;        }    }    return res;}int main(){    //freopen("C:\\Users\\zfh\\Desktop\\in.txt", "r", stdin);    while(scanf("%d", &k) != -1 && k)    {        scanf(" %s", s);        n = 0; m = 300;        len1 = strlen(s);        for(int i = 0; i < len1; i++) str[n++] = s[i];        str[n++] = 0;        scanf(" %s", s);        len2 = strlen(s);        for(int i = 0; i < len2; i++) str[n++] = s[i];        str[n] = 1;        getSa();        getHeight();        int sp = 0, tp = 0;        long long ans = 0;        for(int i = 0; i < n; i++)        {            if(height[i] >= k)            {                tp = i;            }            else            {                ans += cal(sp, tp, 1);                ans += cal(sp, tp, 2);                sp = tp = i;            }        }        ans += cal(sp, tp, 1);        ans += cal(sp, tp, 2);        printf("%I64d\n", ans);    }    return 0;}

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