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题目大意：给你两个字符串，让你求出来两个字符串之间的重复子串长度大于k的有多少个。
解题思路：
先说论文上给的解释：基本思路是计算A的所有后缀和B的所有后缀之间的最长公共前缀的长度，把最长公共前缀长度不小于k的部分全部加起来。先将两个字符串连起来，中间用一个没有出现过的字符隔开。按height值分组后，接下来的工作便是快速的统计每组中后缀之间的最长公共前缀之和。扫描一遍，每遇到一个B的后缀就统计与前面的A的后缀能产生多少个长度不小于k的公共子串，这里A的后缀需要用一个单调的栈来高效的维护。然后对A也这样做一次。具体的细节留给读者思考。
给的解释很模糊，我们自己来补充一下，首先根据LCP定理，LCP(i, j) = min{LCP(k-1, k)|i+1 <= k <= j};所以hieght连续的在一起的如果都小于k了那么这段就可以不要了，这就是分组。分组的同时还要用到单调栈使得每组里面都包含一个单调递增的序列，这样算起来就比较简单省事了啊。注意就是去掉重复的内容，已经加过的不要再重复加了啊。
栈中的每一个元素拥有两个属性，第一个是其值为多少，第二个是前面还有多少个能够提供这个值的一共有多少个（如果新加入的height值比之前较小时，将回收之前的height值，将其视为同一高度，直到遇到比它小的）。需要对height数组作两次。
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 7686 | Accepted: 2549 |
Description
A substring of a string T is defined as:
Given two strings A, B and one integer K, we define S, a set of triples (i, 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
2 aababaa abaabaa 1 xx xx 0
Sample Output
22 5
Source
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)
using namespace std;
inline int read()
{
char ch;
bool flag = false;
int a = 0;
while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
if(ch != '-')
{
a *= 10;
a += ch - '0';
}
else
{
flag = true;
}
while(((ch = getchar()) >= '0') && (ch <= '9'))
{
a *= 10;
a += ch - '0';
}
if(flag)
{
a = -a;
}
return a;
}
void write(int a)
{
if(a < 0)
{
putchar('-');
a = -a;
}
if(a >= 10)
{
write(a / 10);
}
putchar(a % 10 + '0');
}
const int maxn = 200010;
int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];
int cmp(int *r, int a, int b, int l)
{
return r[a] == r[b] && r[a+l] == r[b+l];
}
void da(int *r, int *sa, int n, int m)
{
int i, j, p, *x = wa, *y = wb;
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[x[i] = r[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[x[i]]] = i;
for(j = 1, p = 1; p < n; j <<= 1, m = p)
{
for(p = 0, i = n-j; i < n; i++) y[p++] = i;
for(i = 0; i < n; i++)
if(sa[i] >= j) y[p++] = sa[i]-j;
for(i = 0; i < n; i++) wv[i] = x[y[i]];
for(i = 0; i < m; i++) ws1[i] = 0;
for(i = 0; i < n; i++) ws1[wv[i]]++;
for(i = 1; i < m; i++) ws1[i] += ws1[i-1];
for(i = n-1; i >= 0; i--) sa[--ws1[wv[i]]] = y[i];
for(swap(x, y), p = 1, x[sa[0]] = 0, i = 1; i < n; i++)
x[sa[i]] = cmp(y, sa[i-1], sa[i], j)?p-1:p++;
}
return ;
}
int rank[maxn], height[maxn];
void calheight(int *r, int *sa, int n)
{
int i, j, k = 0;
for(i = 1; i <= n; i++) rank[sa[i]] = i;
for(i = 0; i < n; height[rank[i++]] = k)
for(k?k--:0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
return;
}
char str1[maxn], str2[maxn];
int seq[maxn];
int sta[maxn][2];
void Del(int n, int len1, int len2, int k)
{
LL sum = 0;
int top = 0;
LL ans = 0;
for(int i = 1; i <= n; i++)
{
if(height[i] < k)
{
ans = 0;
top = 0;
continue;
}
int cnt = 0;
if(sa[i-1] < len1)
{
cnt++;
ans += (height[i]-k+1);
}
while(top && sta[top-1][0] >= height[i])
{
top--;
ans -= sta[top][1]*(sta[top][0]-height[i]);
cnt += sta[top][1];
}
sta[top][0] = height[i];
sta[top++][1] = cnt;
if(sa[i] > len1) sum += ans;
}
top = 0;
ans = 0;
for(int i = 1; i <= n; i++)
{
if(height[i] < k)
{
ans = 0;
top = 0;
continue;
}
int cnt = 0;
if(sa[i-1] > len1)
{
cnt++;
ans += (height[i]-k+1);
}
while(top && sta[top-1][0] >= height[i])
{
top--;
ans -= sta[top][1]*(sta[top][0]-height[i]);
cnt += sta[top][1];
}
sta[top][0] = height[i];
sta[top++][1] = cnt;
if(sa[i] < len1) sum += ans;
}
cout<<sum<<endl;
}
int main()
{
int k;
while(~scanf("%d", &k) && k)
{
scanf("%s %s",str1, str2);
int len1 = strlen(str1);
int len2 = strlen(str2);
int ans = 0;
for(int i = 0; i < len1; i++)
seq[ans++] = str1[i];
seq[ans++] = 1;
for(int i = 0; i < len2; i++)
seq[ans++] = str2[i];
seq[ans] = 0;
da(seq, sa, ans+1, 130);
calheight(seq, sa, ans);
Del(ans, len1, len2, k);
}
return 0;
}
POJ 3415 Common Substrings(后缀数组求重复字串)
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原文地址：http://blog.csdn.net/xu12110501127/article/details/43191907
