Common Substrings POJ - 3415(长度不小于k的公共子串的个数)

时间：2024-11-13 22:06:08

题意：

　　给定两个字符串A 和 B， 求长度不小于 k 的公共子串的个数（可以相同）

Common Substrings POJ - 3415(长度不小于k的公共子串的个数)

分两部分求和sa[i-1] > len1  sa[i] < len1  和  sa[i-1] < len1   sa[i] > len1

Common Substrings POJ - 3415(长度不小于k的公共子串的个数)

#include <iostream>

#include <cstdio>

#include <sstream>

#include <cstring>

#include <map>

#include <cctype>

#include <set>

#include <vector>

#include <stack>

#include <queue>

#include <algorithm>

#include <cmath>

#define rap(i, a, n) for(int i=a; i<=n; i++)

#define rep(i, a, n) for(int i=a; i<n; i++)

#define lap(i, a, n) for(int i=n; i>=a; i--)

#define lep(i, a, n) for(int i=n; i>a; i--)

#define rd(a) scanf("%d", &a)

#define rlld(a) scanf("%lld", &a)

#define rc(a) scanf("%c", &a)

#define rs(a) scanf("%s", a)

#define MOD 2018

#define LL long long

#define ULL unsigned long long

#define Pair pair<int, int>

#define mem(a, b) memset(a, b, sizeof(a))

#define _  ios_base::sync_with_stdio(0),cin.tie(0)

//freopen("1.txt", "r", stdin);

using namespace std;

const int maxn = , INF = 0x7fffffff;

int s[maxn];

int sa[maxn], t[maxn], t2[maxn], c[maxn], n;

int ran[maxn], height[maxn];

void get_sa(int m)

{

    int i, *x = t, *y = t2;

    for(i = ; i < m; i++) c[i] = ;

    for(i = ; i < n; i++) c[x[i] = s[i]]++;

    for(i = ; i < m; i++) c[i] += c[i-];

    for(i = n-; i >= ; i--) sa[--c[x[i]]] = i;

    for(int k = ; k <= n; k <<= )

    {

        int p = ;

        for(i = n-k; i < n; i++) y[p++] = i;

        for(i = ; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;

        for(i = ; i < m; i++) c[i] = ;

        for(i = ; i < n; i++) c[x[y[i]]]++;

        for(i = ; i< m; i++) c[i] += c[i-];

        for(i = n-; i >= ; i--) sa[--c[x[y[i]]]] = y[i];

        swap(x, y);

        p = ; x[sa[]] = ;

        for(i = ; i < n; i++)

            x[sa[i]] = y[sa[i-]] == y[sa[i]] && y[sa[i-]+k] == y[sa[i]+k] ? p- : p++;

        if(p >= n) break;

        m = p;

    }

    int k = ;

    for(i = ; i < n; i++) ran[sa[i]] = i;

    for(i = ; i < n; i++)

    {

        if(k) k--;

        int j = sa[ran[i]-];

        while(s[i+k] == s[j+k]) k++;

        height[ran[i]] = k;

    }

}

int k, top, num;

LL sum, ans;

char s1[maxn], s2[maxn];

int stac[maxn], cnt[maxn];

int main()

{

    while(~rd(k) && k)

    {

        top = sum = num = ans = n = ;

        rs(s1); rs(s2);

        int len1 = strlen(s1);

        int len2 = strlen(s2);

        rep(i, , len1)

            s[n++] = s1[i];

        s[n++] = '#';

        rep(i, , len2)

            s[n++] = s2[i];

        s[n++] = ;

        get_sa();

        rep(i, , n)

        {

            if(height[i] < k)

            {

                sum = top = ;

                continue;

            }

            int num = ;

            while(top && height[i] < stac[top])  //维持单调递增栈 可能当前sa[i-1] < len1 但height是连续的 所以短板效应替换栈中元素

            {                                      //而它自己如果sa[i-1] < len1 那么下面的 num是不加1的 即自己不算在内

                sum -= (LL)(stac[top] - k + ) * cnt[top];

                sum += (LL)(height[i] - k + ) * cnt[top];

                num += cnt[top];

                top--;

            }

            stac[++top] = height[i];

            if(sa[i-] > len1)                  //扫描B串

            {

                sum += (LL)(height[i] - k + );

                cnt[top] = num + ;

            }

            else

                cnt[top] = num;

            if(sa[i] < len1)

                ans += sum;

        }

        rep(i, , n)

        {

            if(height[i] < k)

            {

                sum = top = ;

                continue;

            }

            int num = ;

            while(top && height[i] < stac[top])

            {

                sum -= (LL)(stac[top] - k + ) * cnt[top];

                sum += (LL)(height[i] - k + ) * cnt[top];

                num += cnt[top];

                top--;

            }

            stac[++top] = height[i];

            if(sa[i-] < len1)                  //扫描A串

            {

                sum += (LL)(height[i] - k + );

                cnt[top] = num + ;

            }

            else

                cnt[top] = num;

            if(sa[i] > len1)

                ans += sum;

        }

        printf("%lld\n", ans);

    }

    return ;

}

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