poj 3415 Common Substrings

题目链接：http://poj.org/problem?id=3415

题目分类：后缀数组

题意：给出两个串和一个数字k，求两个串的公共字串大于等于k的数目

代码：

//#include<bits/stdc++.h>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<string.h>
 
using namespace std;
 
#define N 200005
#define LL long long
 
int wa[N],wb[N],wm[N],wv[N],sa[N];
int *rank,height[N],s[N],a[N];
//sa:字典序中排第i位的起始位置在str中第sa[i]
//rank:就是str第i个位置的后缀是在字典序排第几
//height：字典序排i和i-1的后缀的最长公共前缀
 
bool cmp(int *r,int a,int b,int l)
{
    return r[a] == r[b] && r[a+l] == r[b+l];
}
 
void getsa(int *r,int *sa,int n,int m)
{
    int *x=wa,*y=wb,*t;
    for(int i=; i<m; ++i)wm[i]=;
    for(int i=; i<n; ++i)wm[x[i]=r[i]]++;
    for(int i=; i<m; ++i)wm[i]+=wm[i-];
    for(int i=n-; i>=; --i)sa[--wm[x[i]]]=i;
    for(int i=,j=,p=; p<n; j=j*,m=p)
    {
        for(p=,i=n-j; i<n; ++i)y[p++]=i;
        for(i=; i<n; ++i)if(sa[i]>=j)y[p++]=sa[i]-j;
        for(i=; i<m; ++i)wm[i]=;
        for(i=; i<n; ++i)wm[x[y[i]]]++;
        for(i=; i<m; ++i)wm[i]+=wm[i-];
        for(i=n-; i>=; --i)sa[--wm[x[y[i]]]]=y[i];
        for(t=x,x=y,y=t,i=p=,x[sa[]]=; i<n; ++i)
        {
            x[sa[i]]=cmp(y,sa[i],sa[i-],j)?p-:p++;
        }
    }
    rank=x;
}
 
void getheight(int *r,int *sa,int n)
{
    for(int i=,j=,k=; i<n; height[rank[i++]]=k)
    {
        for(k?--k:,j=sa[rank[i]-]; r[i+k] == r[j+k]; ++k);
    }
}
int k;
char s1[N];
int len1;
 
LL solve(int n,int len,int k)
{
    int *mark=wa,*sta=wb,top=,i;
    LL sum=,num[]= {};
    for(i = ;i<=n;i++)
    {
        if(height[i]<k)
        {
            top = num[] = num[] =;
        }
        else
        {
            for(int size = top; size&&sta[size]>height[i]-k+; size--)
            {
                num[mark[size]] += height[i]-k+-sta[size];
                sta[size] = height[i]-k+;
            }
            sta[++top] = height[i]-k+;
            if(sa[i-]<len) mark[top] = ;
            if(sa[i-]>len) mark[top] = ;
            num[mark[top]]+=height[i]-k+;
            if(sa[i]<len) sum+=num[];
            if(sa[i]>len) sum+=num[];
        }
    }
    return sum;
}
 
int main()
{
    int i,j;
    while(~scanf("%d",&k),k)
    {
        scanf("%s",s1);
        int n = ;
        for(n = ;s1[n]!='\0';n++)
            s[n] = s1[n];
        s[len1=n] = '#';
        scanf("%s",s1+n+);
        n++;
        for(;s1[n]!='\0';n++)
            s[n] = s1[n];
        s[n] = ;
        getsa(s,sa,n+,);
        getheight(s,sa,n);
        printf("%lld\n",solve(n,len1,k));
    }
    return ;
}