【codechef】Common Strings（后缀数组）

You are given two strings A and B. Find the number of distinct strings which appear in both A and B . A string s is said to appear in S iff s is a substring (appears contiguously) of S.

Input

• The first line of the input contains an integer T denoting the number of test cases. The description ofT test cases follows.
• Each test case consists of two lines.
• The first line contains two space separated integers n1 and n2 denoting the lengths of A and B.
• The second line contains two space separated strings A and B.

Output

• For each test case output a single number denoting the number of distinct strings appearing in Aand B .

Constraints

• 1 ≤ T ≤ 10⁴
• 1 ≤ n1, n2 ≤ 10⁵
• Sum of n1 + n2 over all test cases ≤ 10⁵
• A is a string consisting of n1 lowercase characters ('a'-'z').
• B is a string consisting of n2 lowercase characters ('a'-'z').

Example

Input:23 5aad zaacd4 4abcd lmnoOutput:30

Explanation

Example case 1. The three strings are "a", "d", "aa".

Example case 2. There are no strings that appear in both A and B.

https://www.codechef.com/IOPC2015/problems/IOPC15G/

kuangbin大神的代码。。还没看懂先瞻仰一下。。

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;/**suffix array*倍增算法  O(n*logn)*待排序数组长度为n,放在0~n-1中，在最后面补一个0*da(str ,n+1,sa,rank,height,  ,   );//注意是n+1;*例如：*n   = 8;*num[]   = { 1, 1, 2, 1, 1, 1, 1, 2, $ };注意num最后一位为0，其他大于0*rank[]  = { 4, 6, 8, 1, 2, 3, 5, 7, 0 };rank[0~n-1]为有效值，rank[n]必定为0无效值*sa[]    = { 8, 3, 4, 5, 0, 6, 1, 7, 2 };sa[1~n]为有效值，sa[0]必定为n是无效值*height[]= { 0, 0, 3, 2, 3, 1, 2, 0, 1 };height[2~n]为有效值**/const int MAXN=200010;int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值//待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,//除s[n-1]外的所有s[i]都大于0，r[n-1]=0//函数结束以后结果放在sa数组中bool cmp(int *r,int a,int b,int l){return r[a] == r[b] && r[a+l] == r[b+l];}void da(int str[],int sa[],int rank[],int height[],int n,int m){    n++;    int i, j, p, *x = t1, *y = t2;    //第一轮基数排序，如果s的最大值很大，可改为快速排序    for(i = 0;i < m;i++)c[i] = 0;    for(i = 0;i < n;i++)c[x[i] = str[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(j = 1;j <= n; j <<= 1)    {        p = 0;        //直接利用sa数组排序第二关键字        for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小        for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;        //这样数组y保存的就是按照第二关键字排序的结果        //基数排序第一关键字        for(i = 0; i < m; i++)c[i] = 0;        for(i = 0; i < n; i++)c[x[y[i]]]++;        for(i = 1; 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x数组计算新的x数组        swap(x,y);        p = 1; x[sa[0]] = 0;        for(i = 1;i < n;i++)            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;        if(p >= n)break;        m = p;//下次基数排序的最大值    }    int k = 0;    n--;for(i = 0;i <= n;i++)rank[sa[i]] = i;    for(i = 0;i < n;i++)    {        if(k)k--;        j = sa[rank[i]-1];        while(str[i+k] == str[j+k])k++;        height[rank[i]] = k;    }}int rank[MAXN],height[MAXN];int r[MAXN],sa[MAXN];char str1[MAXN],str2[MAXN];bool check(int i,int j,int n,int m){return (i < n && j > n) || (i > n && j < n);}int main(){    int T;int n,m;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);scanf("%s%s",str1,str2);for(int i = 0;i < n;i++)r[i] = str1[i]-'a'+1;r[n] = 27;for(int i = 0;i < m;i++)r[n+1+i] = str2[i]-'a'+1;r[n+m+1] = 0;da(r,sa,rank,height,n+m+1,28);long long ans = 0;int tmp = 0;for(int i = 2;i <= n+m+1;i++){tmp = min(tmp,height[i]);if(check(sa[i],sa[i-1],n,m)){ans += height[i]-tmp;tmp = height[i];}}cout<<ans<<endl;}    return 0;}