UVa10526 - Intellectual Property(后缀数组的应用)

题意：给出两个字符串s1,s2,求s2在s1中的出现的子串，输出其起始位置和长度，按长度从大到小，位置从小到大排序输出，忽略重叠的子串

代码如下：

#include <iostream>
#include <string>
#include <fstream>
#include <cstring>
#include <vector>
#include <algorithm>using namespace std;const int N = 256;
const int MIN = 0x3f3f3f3f;
const int MAXN = 100010;class SuffixArray
{
public:SuffixArray(int* pa, int n);~SuffixArray();void buildSa();void getHeight();const int* getSa();const int* getH();
private:int *m_prank;int *m_pheight;int *m_psa;int m_n;int* m_a;
};SuffixArray::SuffixArray(int* pa, int n):m_prank(NULL), m_pheight(NULL),  m_psa(NULL)
{m_a = pa;m_n = n;
}SuffixArray::~SuffixArray()
{delete []m_prank;delete []m_psa;delete []m_pheight;
}void SuffixArray::buildSa()
{m_prank = new int[m_n];int n = max(N, m_n);int* tmp = new int[m_n];int* c = new int[n];m_psa = new int[m_n];m_pheight = new int[m_n];memset(c, 0, sizeof(int) * n);for (int i = 0; i < m_n; i++){m_prank[i] = m_a[i];c[m_prank[i]]++;}for (int i = 1; i < n; i++){c[i] += c[i - 1];}for (int i = m_n - 1; i >= 0; i--){m_psa[--c[m_prank[i]]] = i;}int m =  m_n;for (int i = 1; i <= m_n; i <<= 1) {int p = 0;for (int j = m_n - i; j < m_n; j++){tmp[p++] = j;}for (int j = 0; j < m_n; j++){if (m_psa[j] >= i){tmp[p++] = m_psa[j] - i;}}memset(c, 0, sizeof(int) * n);for (int j = 0; j < m_n; j++){c[m_prank[tmp[j]]]++;}for (int j = 1; j < n; j++){c[j] += c[j - 1];}for (int j = m_n - 1; j >= 0; j--){m_psa[--c[m_prank[tmp[j]]]] = tmp[j];}swap(m_prank, tmp);m_prank[m_psa[0]] = 0;m = 0;for (int j = 1; j < m_n; j++){if (tmp[m_psa[j - 1]] == tmp[m_psa[j]] && m_psa[j- 1] + i < m_n && m_psa[j] + i < m_n && tmp[m_psa[j - 1] + i] == tmp[m_psa[j] + i]){m_prank[m_psa[j]] = m;}else {m_prank[m_psa[j]] = ++m;}}if (m >= m_n - 1) break;}delete []c;delete []tmp;return;
}void SuffixArray::getHeight()
{m_pheight[0] = 0;int k = 0;for (int i = 0; i < m_n; i++){int curPos = m_prank[i];if (curPos == 0) continue;curPos--;int j = m_psa[curPos];if (k){k--;}while (i + k < m_n && j + k < m_n && m_a[i + k] == m_a[j + k]) k++;m_pheight[m_prank[i]] = k;}
}const int* SuffixArray::getSa()
{return m_psa;
}const int* SuffixArray::getH()
{return m_pheight;
}class Result
{
public:int len, pos;bool operator < (const Result& other) const{if (len != other.len){return len > other.len;}return pos < other.pos;}
};int a[MAXN];int main(int argc, char **argv)
{
#ifndef ONLINE_JUDGEifstream fin("f:\\OJ\\uva_in.txt");streambuf* old = cin.rdbuf(fin.rdbuf());
#endifint k;int testCase = 0;while (cin >> k){if (0 == k){break;}string tdpCode, jcnCode, tmp;getline(cin, tmp);getline(cin, tmp);while (getline(cin, tmp)){if ("END TDP CODEBASE" == tmp){break;}tdpCode += tmp;tdpCode += "\n";}getline(cin, tmp);while (getline(cin, tmp)){if ("END JCN CODEBASE" == tmp){break;}jcnCode += tmp;jcnCode += "\n";}string str = jcnCode + tdpCode;int tdpCodeLen = tdpCode.length();for (int i = 0; i < tdpCodeLen; i++){a[i] = tdpCode[i];}a[tdpCodeLen] = N - 1;int jcnCodeLen = jcnCode.length();for (int i = 0; i < jcnCodeLen + 1; i++){a[tdpCodeLen + i + 1] = jcnCode[i];}SuffixArray suffixArray(a, tdpCodeLen + jcnCodeLen + 1);suffixArray.buildSa();suffixArray.getHeight();const int* psa = suffixArray.getSa();const int* pheight = suffixArray.getH();int saLen = tdpCodeLen + jcnCodeLen + 1;int* pCnt = new int[jcnCodeLen];memset(pCnt, 0, sizeof(int) * jcnCodeLen);int minVal = -1;for (int i = 0; i < saLen; i++){if (psa[i] < tdpCodeLen){minVal = MIN;}else if (psa[i] > tdpCodeLen){if (-1 == minVal){continue;}minVal = min(pheight[i], minVal);pCnt[psa[i] - 1 - tdpCodeLen] = max(pCnt[psa[i] - 1 - tdpCodeLen], minVal);}}minVal = -1;for (int i = saLen - 1; i >= 0; i--){if (psa[i] < tdpCodeLen){minVal = MIN;}else if (psa[i] > tdpCodeLen){if (-1 == minVal){continue;}minVal = min(pheight[i + 1], minVal);pCnt[psa[i] - 1 - tdpCodeLen] = max(pCnt[psa[i] - 1 - tdpCodeLen], minVal);}}vector<Result> result;int curPos = -1;for (int i = 0; i < jcnCodeLen; i++){if ((i + pCnt[i] <= curPos) || pCnt[i] <= 0) continue;result.push_back((Result){pCnt[i], i});curPos = i + pCnt[i];}delete []pCnt;sort(result.begin(), result.end());if (testCase){cout << endl;}cout << "CASE " << ++testCase << endl;for (int i = 0; i < min(k, (int)result.size()); i++){cout << "INFRINGING SEGMENT " << i + 1 << " LENGTH " << result[i].len << " POSITION " << result[i].pos << endl;cout << jcnCode.substr(result[i].pos, result[i].len) << endl;}}#ifndef ONLINE_JUDGEcin.rdbuf(old);
#endif
}

UVa10526 - Intellectual Property(后缀数组的应用)相关推荐

UVA 10526 - Intellectual Property (后缀数组)
UVA 10526 - Intellectual Property 题目链接题意:给定两个问题,要求找出第二个文本抄袭第一个文本的所有位置和长度,输出前k个,按长度从大到小先排,长度一样的按位置从小 ...
寻找一个字符串的重复子串后缀数组
什么是后缀数组令字符串 S=S[1]S[2]...S[n]S=S[1]S[2]...S[n]{\displaystyle S=S[1]S[2]...S[n]} , S[i,j]S[i,j]{\dis ...
【2012百度之星/资格赛】H:用户请求中的品牌 [后缀数组]
时间限制: 1000ms 内存限制: 65536kB 描述馅饼同学是一个在百度工作,做用户请求(query)分析的同学,他在用户请求中经常会遇到一些很奇葩的词汇.在比方说"johnsonj ...
Boring counting HDU - 3518 （后缀数组）
Boring counting \[ Time Limit: 1000 ms \quad Memory Limit: 32768 kB \] 题意给出一个字符串,求出其中出现两次及以上的子串个数,要 ...
HDU4080 Stammering Aliens（二分 + 后缀数组）
题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=4080 Description Dr. Ellie Arroway has establish ...
后缀数组 + Hash + 二分 or Hash + 二分 + 双指针求 LCP ---- 2017icpc 青岛 J Suffix (假题！！)
题目链接题目大意: 就是给你n个串每个串取一个后缀,要求把串拼起来要求字典序最小!! sum_length_of_n≤5e5sum\_length\_of\_n\leq 5e5sum_length_ ...
后缀数组 ---- 2018~2019icpc焦作H题[后缀数组+st表+二分+单调栈]
题目链接题目大意: 给出nnn个数,定义f[l,r]f[l,r]f[l,r]表示区间[l,r][l,r][l,r]的最大值,求所有子区间的最大值的和,要求相同的子区间只能算一次比如数列 5 6 ...
[Ahoi2013]差异[后缀数组+单调栈]
链接解题思路:很明显前面∑1<=i<j<=nlen(Ti)+len(Tj)\sum_{1<=i<j<=n}len(T_i)+len(T_j)∑1<=i< ...
poj2217详解（后缀数组 + 高度数组）
题目大概意思就是给两个字符串,求最长公共字符串子串长度我们可以考虑用后缀数组和高度数组一个字符串中最长的两个相同字符串长度, 不就是后缀数组中相邻两个后缀的最长公共前缀, 不就是高度数组 ...

UVa10526 - Intellectual Property(后缀数组的应用)

UVa10526 - Intellectual Property(后缀数组的应用)相关推荐

最新文章

热门文章