求N个字符串最长的公共子串。这题数据比较水,暴力第一个字符串的子串也可以过。
初学后缀数组,有很多不明白的东西,此题后缀数组的代码在网上也是一把抓。
说实话我确实还不懂后缀数组,但是后缀数组太强大了,只能硬着头皮照着葫芦画瓢了。
贴下代码方便以后查阅吧。。。
感觉后缀数组的应用最主要的还是height数组,看懂倍增算法排序后缀已经非常困难了。
然后再理解height数组怎么用也不是一件容易的事情。然后貌似height数组最关键的用法是
枚举某一个长度的子串时候,比如长度为k,能够用这个k对height数组进行分组,这个罗穗骞
的论文里面有个求不重叠最长重复子串的例子说明了这个height数组分组的思路,不过我现在
还是不怎么理解。。。
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int MAX_N = 110;
const int MAX_L = MAX_N * MAX_N;
char szStr[MAX_N];
int nNum[MAX_L];
int nLoc[MAX_L];
bool bVisit[MAX_N];
int sa[MAX_L], rank[MAX_L], height[MAX_L];
int wa[MAX_L], wb[MAX_L], wv[MAX_L], wd[MAX_L];
int cmp(int* r, int a, int b, int l)
{
return r[a] == r[b] && r[a + l] == r[b + l];
}
//倍增算法,r为待匹配数组,n为总长度,m为字符串范围
void da(int* r, int n, int m)
{
int i, j, p, *x = wa, *y = wb;
for (i = 0; i < m; ++i) wd[i] = 0;
for (i = 0; i < n; ++i) wd[x[i] = r[i]]++;
for (i = 1; i < m; ++i) wd[i] += wd[i - 1];
for (i = n - 1; i >= 0; --i) sa[--wd[x[i]]] = i;
for (j = 1, p = 1; p < n; j *= 2, 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) wd[i] = 0;
for (i = 0; i < n; ++i) wd[wv[i]]++;
for (i = 1; i < m; ++i) wd[i] += wd[i - 1];
for (i = n - 1; i >= 0; --i) sa[--wd[wv[i]]] = y[i];
swap(x, y);
for (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++;
}
}
}
//求height数组
void calHeight(int* r, 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)
{
if (k) --k;
for(j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k++);
}
}
bool Check(int nMid, int nLen, int nN)
{
int nCnt = 0;
memset(bVisit, false, sizeof(bVisit));
for (int i = 2; i <= nLen; ++i)
{
if (nMid > height[i])
{
nCnt = 0;
memset(bVisit, false, sizeof(bVisit));
continue;
}
if (!bVisit[nLoc[sa[i - 1]]])
{
bVisit[nLoc[sa[i - 1]]] = true;
++nCnt;
}
if (!bVisit[nLoc[sa[i]]])
{
bVisit[nLoc[sa[i]]] = true;
++nCnt;
}
if (nCnt == nN) return true;
}
return false;
}
int main()
{
int nT;
scanf("%d", &nT);
while (nT--)
{
int nN;
int nEnd = 300;
int nP = 0;
scanf("%d", &nN);
for (int i = 1; i <= nN; ++i)
{
scanf("%s", szStr);
char* pszStr;
for (pszStr = szStr; *pszStr; ++pszStr)
{
nLoc[nP] = i;
nNum[nP++] = *pszStr;
}
nLoc[nP] = nEnd;
nNum[nP++] = nEnd++;
reverse(szStr, szStr + strlen(szStr));
for (pszStr = szStr; *pszStr; ++pszStr)
{
nLoc[nP] = i;
nNum[nP++] = *pszStr;
}
nLoc[nP] = nEnd;
nNum[nP++] = nEnd++;
}
nNum[nP] = 0;
da(nNum, nP + 1, nEnd);
calHeight(nNum, nP);
int nLeft = 1, nRight = strlen(szStr), nMid;
int nAns = 0;
while (nLeft <= nRight)
{
nMid = (nLeft + nRight) / 2;
if (Check(nMid, nP, nN))
{
nLeft = nMid + 1;
nAns = nMid;
}
else nRight = nMid - 1;
}
printf("%d\n", nAns);
}
return 0;
}