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| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 1879 | Accepted: 568 |
Description
Minifacer was very happy these days because he has learned the algorithm of KMP recently. Yet his elder brother, Hugefacer, thought that Minifacer needs a deeper understanding of this algorithm. Thus Hugefacer decided to play a game with his little brother to enhance his skills.
First, Hugefacer wrote down two strings S1 and S2. Then Minifacer tried to find a substring S3 of S1 which meets the following requirements: 1) S3 should have a length of k (which is a constant value); 2) S3 should also be the substring of S2. After several rounds, Hugefacer found that this game was too easy for his clever little brother, so he added another requirement: 3) the extended string of S3 should NOT be the substring of S2. Here the extended string of S3 is defined as S3 plus its succeed character in S1 (if S3 does not have a succeed character in S1, the extended string of S3 is S3 + ‘ ‘ which will never appear in S2). For example, let S1 be "ababc", if we select the substring from the first character to the second character as S3 (so S3 equals "ab"), its extended string should be "aba"; if we select the substring from the third character to the fourth character as S3, its extended string should be "abc"; if we select the substring from the fourth character to the fifth character as S3, its extended string should be "bc".
Since the difficult level of the game has been greatly increased after the third requirement was added, Minifacer was not able to win the game and he thought that maybe none of the substring would meet all the requirements. In order to prove that Minifacer was wrong, Hugefacer would like to write a program to compute number of substrings that meet the three demands (Note that two strings with same appearance but different positions in original string S1 should be count twice). Since Hugefacer do not like characters, he will use non-negative integers (range from 0 to 10000) instead.
Input
There are multiple test cases. Each case contains three lines: the first line contains three integers n, m and k where n represents the length of S1, m represents the length of S2 and k represents the length of substring; the second line contains string S1 and the third line contains string S2. Here 0 ≤ n, m ≤ 50000. Input ends with EOF.
Output
For each test case, output a number in a line stand for the total number of substrings that meet the three requirements.
Sample Input
5 5 2 1 2 1 2 3 1 2 3 4 5 5 5 3 1 2 1 2 3 1 2 3 4 5
Sample Output
2 1
Source
POJ Monthly Contest – 2009.04.05, Facer
就是求a串的子串有几个能和b串的所有子串的LCP==k
看到题除了用SA+RMQ暴力一下LCP没啥思路，，这样肯定是超时，于是借助强大的百度，，参考了一下别人的代码，很巧妙，线性扫描，因为求的恰好是k，可以求大于等于k的再减去大于等于k+1的，把俩串组合到一起后，用height数组把后缀分组，看看每组中有几个是第一个串的几个是第二个串的，如果都是第一个串的就不行了，不是的话，加上第一个串的数目
ac代码
Problem: 3729 User: kxh1995 Memory: 2512K Time: 657MS Language: C++ Result: Accepted
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
using namespace std;
#define INF 0x3f3f3f3f
char str[103030];
int sa[103030],Rank[103030],rank2[103030],height[103030],c[103030],*x,*y,s[100030];
void cmp(int n,int sz)
{
int i;
memset(c,0,sizeof(c));
for(i=0;i<n;i++)
c[x[y[i]]]++;
for(i=1;i<sz;i++)
c[i]+=c[i-1];
for(i=n-1;i>=0;i--)
sa[--c[x[y[i]]]]=y[i];
}
void build_sa(int *s,int n,int sz)
{
x=Rank,y=rank2;
int i,j;
for(i=0;i<n;i++)
x[i]=s[i],y[i]=i;
cmp(n,sz);
int len;
for(len=1;len<n;len<<=1)
{
int yid=0;
for(i=n-len;i<n;i++)
{
y[yid++]=i;
}
for(i=0;i<n;i++)
if(sa[i]>=len)
y[yid++]=sa[i]-len;
cmp(n,sz);
swap(x,y);
x[sa[0]]=yid=0;
for(i=1;i<n;i++)
{
if(y[sa[i-1]]==y[sa[i]]&&sa[i-1]+len<n&&sa[i]+len<n&&y[sa[i-1]+len]==y[sa[i]+len])
x[sa[i]]=yid;
else
x[sa[i]]=++yid;
}
sz=yid+1;
if(sz>=n)
break;
}
for(i=0;i<n;i++)
Rank[i]=x[i];
}
void getHeight(int *s,int n)
{
int k=0;
for(int i=0;i<n;i++)
{
if(Rank[i]==0)
continue;
k=max(0,k-1);
int j=sa[Rank[i]-1];
while(s[i+k]==s[j+k])
k++;
height[Rank[i]]=k;
}
}
int n,m,k,len;
__int64 fun(int k)
{
int l,r,i;
l=0;
r=0;
__int64 ans=0;
for(i=1;i<=len;i++)
{
if(height[i]<k)
{
if(r)
ans+=(__int64)l;
l=r=0;
if(sa[i]<n)
l++;
else
r++;
}
else
{
if(sa[i]<n)
l++;
else
r++;
}
}
return ans;
}
int main()
{
while(scanf("%d%d%d",&n,&m,&k)!=EOF)
{
int i;
for(i=0;i<n;i++)
{
scanf("%d",&s[i]);
s[i]++;
}
s[n]=10002;
len=n+m+1;
for(i=n+1;i<len;i++)
{
scanf("%d",&s[i]);
s[i]++;
}
s[len]=0;
build_sa(s,len+1,10003);
getHeight(s,len);
printf("%I64d\n",fun(k)-fun(k+1));
}
}
版权声明：本文为博主原创文章，未经博主允许不得转载。
POJ题目3229 Facer’s string（后缀数组求a串长度为k子串有几个出现在b串）
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原文地址：http://blog.csdn.net/yu_ch_sh/article/details/48089631
