Diff
checker
文本
文本
图像
文档
Excel
文件夹
Legal
Enterprise
桌面版
定价
登录
下载 Diffchecker 桌面版
比较文本
查找两个文本文件之间的差异
工具
历史
显示
视图
拆分
统一
实时编辑器
隐藏空白更改
折叠未更改行
关闭换行
比对精度
智能
语法
无
更改外观
处理
忽略
文本转换
转到第一个差异
编辑输入
Diffchecker Desktop
运行Diffchecker最安全的方式。获取Diffchecker桌面应用:您的差异永远不会离开您的电脑!
获取桌面版
Stupid
创建于
6年前
差异永不过期
清除
导出
分享
解释
1 删除
行
总计
删除
字符
总计
删除
要继续使用此功能,请升级到
Diff
checker
Pro
查看价格
182 行
全部复制
1 添加
行
总计
添加
字符
总计
添加
要继续使用此功能,请升级到
Diff
checker
Pro
查看价格
182 行
全部复制
#include<bits/stdc++.h>
#include<bits/stdc++.h>
using namespace std;
using namespace std;
#include "spoj.h"
#include "spoj.h"
const int N = 1e5 + 9;
const int N = 1e5 + 9;
/*
/*
-> diff(v) = len(v) - len(link(v))
-> diff(v) = len(v) - len(link(v))
-> series link will lead from the vertex v to the vertex u corresponding
-> series link will lead from the vertex v to the vertex u corresponding
to the maximum suffix palindrome of v which satisfies diff(v) != diff(u)
to the maximum suffix palindrome of v which satisfies diff(v) != diff(u)
-> path within series links to the root contains only O(log n) vertices
-> path within series links to the root contains only O(log n) vertices
-> cnt contains the number of palindromic suffixes of the node
-> cnt contains the number of palindromic suffixes of the node
*/
*/
struct PalindromicTree {
struct PalindromicTree {
struct node {
struct node {
int nxt[26], len, st, en, link, diff, slink, cnt, oc;
int nxt[26], len, st, en, link, diff, slink, cnt, oc;
};
};
string s; vector<node> t;
string s; vector<node> t;
int sz, last;
int sz, last;
PalindromicTree() {}
PalindromicTree() {}
PalindromicTree(string _s) {
PalindromicTree(string _s) {
s = _s; int n = s.size();
s = _s; int n = s.size();
t.clear(); t.resize(n + 9);
t.clear(); t.resize(n + 9);
sz = 2, last = 2;
sz = 2, last = 2;
t[1].len = -1, t[1].link = 1;
t[1].len = -1, t[1].link = 1;
t[2].len = 0, t[2].link = 1;
t[2].len = 0, t[2].link = 1;
t[1].diff = t[2].diff = 0;
t[1].diff = t[2].diff = 0;
t[1].slink = 1; t[2].slink = 2;
t[1].slink = 1; t[2].slink = 2;
}
}
int extend(int pos) { // returns 1 if it creates a new palindrome
int extend(int pos) { // returns 1 if it creates a new palindrome
int cur = last, curlen = 0;
int cur = last, curlen = 0;
int ch = s[pos] - 'a';
int ch = s[pos] - 'a';
while (1) {
while (1) {
curlen = t[cur].len;
curlen = t[cur].len;
if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) break;
if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) break;
cur = t[cur].link;
cur = t[cur].link;
}
}
if (t[cur].nxt[ch]) {
if (t[cur].nxt[ch]) {
last = t[cur].nxt[ch];
last = t[cur].nxt[ch];
t[last].oc++;
t[last].oc++;
return 0;
return 0;
}
}
sz++; last = sz;
sz++; last = sz;
t[sz].oc = 1; t[sz].len = t[cur].len + 2;
t[sz].oc = 1; t[sz].len = t[cur].len + 2;
t[cur].nxt[ch] = sz;
t[cur].nxt[ch] = sz;
t[sz].en = pos; t[sz].st = pos - t[sz].len + 1;
t[sz].en = pos; t[sz].st = pos - t[sz].len + 1;
if (t[sz].len == 1) {
if (t[sz].len == 1) {
t[sz].link = 2; t[sz].cnt = 1;
t[sz].link = 2; t[sz].cnt = 1;
t[sz].diff = 1; t[sz].slink = 2;
t[sz].diff = 1; t[sz].slink = 2;
return 1;
return 1;
}
}
while (1) {
while (1) {
cur = t[cur].link; curlen = t[cur].len;
cur = t[cur].link; curlen = t[cur].len;
if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) {
if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) {
t[sz].link = t[cur].nxt[ch];
t[sz].link = t[cur].nxt[ch];
break;
break;
}
}
}
}
t[sz].cnt = 1 + t[t[sz].link].cnt;
t[sz].cnt = 1 + t[t[sz].link].cnt;
t[sz].diff = t[sz].len - t[t[sz].link].len;
t[sz].diff = t[sz].len - t[t[sz].link].len;
if (t[sz].diff == t[t[sz].link].diff) t[sz].slink = t[t[sz].link].slink;
if (t[sz].diff == t[t[sz].link].diff) t[sz].slink = t[t[sz].link].slink;
else t[sz].slink = t[sz].link;
else t[sz].slink = t[sz].link;
return 1;
return 1;
}
}
void calc_occurrences() {
void calc_occurrences() {
for (int i = sz; i >= 3; i--) t[t[i].link].oc += t[i].oc;
for (int i = sz; i >= 3; i--) t[t[i].link].oc += t[i].oc;
}
}
};
};
// len -> largest string length of the corresponding endpos-equivalent class
// len -> largest string length of the corresponding endpos-equivalent class
// link -> longest suffix that is another endpos-equivalent class.
// link -> longest suffix that is another endpos-equivalent class.
// firstpos -> 1 indexed end position of the first occurrence of the largest string of that node
// firstpos -> 1 indexed end position of the first occurrence of the largest string of that node
// minlen(v) -> smallest string of node v = len(link(v)) + 1
// minlen(v) -> smallest string of node v = len(link(v)) + 1
// terminal nodes -> store the suffixes
// terminal nodes -> store the suffixes
struct SuffixAutomaton {
struct SuffixAutomaton {
struct node {
struct node {
int len, link, firstpos;
int len, link, firstpos;
map<char, int> nxt;
map<char, int> nxt;
};
};
int sz, last;
int sz, last;
vector<node> t;
vector<node> t;
vector<int> terminal;
vector<int> terminal;
vector<int> dp;
vector<int> dp;
vector<vector<int>> g;
vector<vector<int>> g;
SuffixAutomaton() {}
SuffixAutomaton() {}
SuffixAutomaton(int n) {
SuffixAutomaton(int n) {
t.resize(2 * n); terminal.resize(2 * n, 0);
t.resize(2 * n); terminal.resize(2 * n, 0);
dp.resize(2 * n, -1); sz = 1; last = 0;
dp.resize(2 * n, -1); sz = 1; last = 0;
g.resize(2 * n);
g.resize(2 * n);
t[0].len = 0; t[0].link = -1; t[0].firstpos = 0;
t[0].len = 0; t[0].link = -1; t[0].firstpos = 0;
}
}
void extend(char c) {
void extend(char c) {
int p = last;
int p = last;
if (t[p].nxt.count(c)) {
if (t[p].nxt.count(c)) {
int q = t[p].nxt[c];
int q = t[p].nxt[c];
if (t[q].len == t[p].len + 1) {
if (t[q].len == t[p].len + 1) {
last = q;
last = q;
return;
return;
}
}
int clone = sz++;
int clone = sz++;
已保存差异
原始文本
打开文件
#include<bits/stdc++.h> using namespace std; #include "spoj.h" const int N = 1e5 + 9; /* -> diff(v) = len(v) - len(link(v)) -> series link will lead from the vertex v to the vertex u corresponding to the maximum suffix palindrome of v which satisfies diff(v) != diff(u) -> path within series links to the root contains only O(log n) vertices -> cnt contains the number of palindromic suffixes of the node */ struct PalindromicTree { struct node { int nxt[26], len, st, en, link, diff, slink, cnt, oc; }; string s; vector<node> t; int sz, last; PalindromicTree() {} PalindromicTree(string _s) { s = _s; int n = s.size(); t.clear(); t.resize(n + 9); sz = 2, last = 2; t[1].len = -1, t[1].link = 1; t[2].len = 0, t[2].link = 1; t[1].diff = t[2].diff = 0; t[1].slink = 1; t[2].slink = 2; } int extend(int pos) { // returns 1 if it creates a new palindrome int cur = last, curlen = 0; int ch = s[pos] - 'a'; while (1) { curlen = t[cur].len; if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) break; cur = t[cur].link; } if (t[cur].nxt[ch]) { last = t[cur].nxt[ch]; t[last].oc++; return 0; } sz++; last = sz; t[sz].oc = 1; t[sz].len = t[cur].len + 2; t[cur].nxt[ch] = sz; t[sz].en = pos; t[sz].st = pos - t[sz].len + 1; if (t[sz].len == 1) { t[sz].link = 2; t[sz].cnt = 1; t[sz].diff = 1; t[sz].slink = 2; return 1; } while (1) { cur = t[cur].link; curlen = t[cur].len; if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) { t[sz].link = t[cur].nxt[ch]; break; } } t[sz].cnt = 1 + t[t[sz].link].cnt; t[sz].diff = t[sz].len - t[t[sz].link].len; if (t[sz].diff == t[t[sz].link].diff) t[sz].slink = t[t[sz].link].slink; else t[sz].slink = t[sz].link; return 1; } void calc_occurrences() { for (int i = sz; i >= 3; i--) t[t[i].link].oc += t[i].oc; } }; // len -> largest string length of the corresponding endpos-equivalent class // link -> longest suffix that is another endpos-equivalent class. // firstpos -> 1 indexed end position of the first occurrence of the largest string of that node // minlen(v) -> smallest string of node v = len(link(v)) + 1 // terminal nodes -> store the suffixes struct SuffixAutomaton { struct node { int len, link, firstpos; map<char, int> nxt; }; int sz, last; vector<node> t; vector<int> terminal; vector<int> dp; vector<vector<int>> g; SuffixAutomaton() {} SuffixAutomaton(int n) { t.resize(2 * n); terminal.resize(2 * n, 0); dp.resize(2 * n, -1); sz = 1; last = 0; g.resize(2 * n); t[0].len = 0; t[0].link = -1; t[0].firstpos = 0; } void extend(char c) { int p = last; if (t[p].nxt.count(c)) { int q = t[p].nxt[c]; if (t[q].len == t[p].len + 1) { last = q; return; } int clone = sz++; t[clone] = t[q]; t[clone].len = t[p].len + 1; t[q].link = clone; last = clone; while (p != -1 && t[p].nxt[c] == q) { t[p].nxt[c] = clone; p = t[p].link; } return; } int cur = sz++; t[cur].len = t[last].len + 1; t[cur].firstpos = t[cur].len; p = last; while (p != -1 && !t[p].nxt.count(c)) { t[p].nxt[c] = cur; p = t[p].link; } if (p == -1) t[cur].link = 0; else { int q = t[p].nxt[c]; if (t[p].len + 1 == t[q].len) t[cur].link = q; else { int clone = sz++; t[clone] = t[q]; t[clone].len = t[p].len + 1; while (p != -1 && t[p].nxt[c] == q) { t[p].nxt[c] = clone; p = t[p].link; } t[q].link = t[cur].link = clone; } } last = cur; } void build(string &s) { for (auto x: s) { extend(x); terminal[last] = 1; } } }; long long distinct(string s) { SuffixAutomaton sa(s.size()); sa.build(s); long long ans = 0; for (int i = 1; i < sa.sz; i++) ans += sa.t[i].len - sa.t[sa.t[i].link].len; return ans; } long long palindromic(string s) { PalindromicTree t(s); for (int i = 0; i < s.size(); i++) t.extend(i); t.calc_occurrences(); long long ans = 0; for (int i = 3; i <= t.sz; i++) ans += t.t[i].cnt; return ans; } char s[N]; int main(){ int t, n; spoj_init(); fscanf(spoj_p_in, "%d", &t); for (int tt = 1; tt <= t; tt++) { fscanf(spoj_p_in, "%d", &n); int z; spoj_assert(fscanf(spoj_t_out, "%d", &z) == 1); spoj_assert(1 <= z && z <= 7 * n); spoj_assert(fscanf(spoj_t_out, "%s", s) == 1); string p = string(s); int sz = p.size(); spoj_assert(z == sz); for (char c: p) spoj_assert('a' <= c && c <= 'z'); long long tot = distinct(p) - palindromic(p); spoj_assert(tot == n); } char extra[100]; spoj_assert(fscanf(spoj_t_out, "%s", extra) != 1); return 0; }
更改后文本
打开文件
#include<bits/stdc++.h> using namespace std; #include "spoj.h" const int N = 1e5 + 9; /* -> diff(v) = len(v) - len(link(v)) -> series link will lead from the vertex v to the vertex u corresponding to the maximum suffix palindrome of v which satisfies diff(v) != diff(u) -> path within series links to the root contains only O(log n) vertices -> cnt contains the number of palindromic suffixes of the node */ struct PalindromicTree { struct node { int nxt[26], len, st, en, link, diff, slink, cnt, oc; }; string s; vector<node> t; int sz, last; PalindromicTree() {} PalindromicTree(string _s) { s = _s; int n = s.size(); t.clear(); t.resize(n + 9); sz = 2, last = 2; t[1].len = -1, t[1].link = 1; t[2].len = 0, t[2].link = 1; t[1].diff = t[2].diff = 0; t[1].slink = 1; t[2].slink = 2; } int extend(int pos) { // returns 1 if it creates a new palindrome int cur = last, curlen = 0; int ch = s[pos] - 'a'; while (1) { curlen = t[cur].len; if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) break; cur = t[cur].link; } if (t[cur].nxt[ch]) { last = t[cur].nxt[ch]; t[last].oc++; return 0; } sz++; last = sz; t[sz].oc = 1; t[sz].len = t[cur].len + 2; t[cur].nxt[ch] = sz; t[sz].en = pos; t[sz].st = pos - t[sz].len + 1; if (t[sz].len == 1) { t[sz].link = 2; t[sz].cnt = 1; t[sz].diff = 1; t[sz].slink = 2; return 1; } while (1) { cur = t[cur].link; curlen = t[cur].len; if (pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos]) { t[sz].link = t[cur].nxt[ch]; break; } } t[sz].cnt = 1 + t[t[sz].link].cnt; t[sz].diff = t[sz].len - t[t[sz].link].len; if (t[sz].diff == t[t[sz].link].diff) t[sz].slink = t[t[sz].link].slink; else t[sz].slink = t[sz].link; return 1; } void calc_occurrences() { for (int i = sz; i >= 3; i--) t[t[i].link].oc += t[i].oc; } }; // len -> largest string length of the corresponding endpos-equivalent class // link -> longest suffix that is another endpos-equivalent class. // firstpos -> 1 indexed end position of the first occurrence of the largest string of that node // minlen(v) -> smallest string of node v = len(link(v)) + 1 // terminal nodes -> store the suffixes struct SuffixAutomaton { struct node { int len, link, firstpos; map<char, int> nxt; }; int sz, last; vector<node> t; vector<int> terminal; vector<int> dp; vector<vector<int>> g; SuffixAutomaton() {} SuffixAutomaton(int n) { t.resize(2 * n); terminal.resize(2 * n, 0); dp.resize(2 * n, -1); sz = 1; last = 0; g.resize(2 * n); t[0].len = 0; t[0].link = -1; t[0].firstpos = 0; } void extend(char c) { int p = last; if (t[p].nxt.count(c)) { int q = t[p].nxt[c]; if (t[q].len == t[p].len + 1) { last = q; return; } int clone = sz++; t[clone] = t[q]; t[clone].len = t[p].len + 1; t[q].link = clone; last = clone; while (p != -1 && t[p].nxt[c] == q) { t[p].nxt[c] = clone; p = t[p].link; } return; } int cur = sz++; t[cur].len = t[last].len + 1; t[cur].firstpos = t[cur].len; p = last; while (p != -1 && !t[p].nxt.count(c)) { t[p].nxt[c] = cur; p = t[p].link; } if (p == -1) t[cur].link = 0; else { int q = t[p].nxt[c]; if (t[p].len + 1 == t[q].len) t[cur].link = q; else { int clone = sz++; t[clone] = t[q]; t[clone].len = t[p].len + 1; while (p != -1 && t[p].nxt[c] == q) { t[p].nxt[c] = clone; p = t[p].link; } t[q].link = t[cur].link = clone; } } last = cur; } void build(string &s) { for (auto x: s) { extend(x); terminal[last] = 1; } } }; long long distinct(string s) { SuffixAutomaton sa(s.size()); sa.build(s); long long ans = 0; for (int i = 1; i < sa.sz; i++) ans += sa.t[i].len - sa.t[sa.t[i].link].len; return ans; } long long palindromic(string s) { PalindromicTree t(s); for (int i = 0; i < s.size(); i++) t.extend(i); t.calc_occurrences(); long long ans = 0; for (int i = 3; i <= t.sz; i++) ans += t.t[i].oc; return ans; } char s[N]; int main(){ int t, n; spoj_init(); fscanf(spoj_p_in, "%d", &t); for (int tt = 1; tt <= t; tt++) { fscanf(spoj_p_in, "%d", &n); int z; spoj_assert(fscanf(spoj_t_out, "%d", &z) == 1); spoj_assert(1 <= z && z <= 7 * n); spoj_assert(fscanf(spoj_t_out, "%s", s) == 1); string p = string(s); int sz = p.size(); spoj_assert(z == sz); for (char c: p) spoj_assert('a' <= c && c <= 'z'); long long tot = distinct(p) - palindromic(p); spoj_assert(tot == n); } char extra[100]; spoj_assert(fscanf(spoj_t_out, "%s", extra) != 1); return 0; }
查找差异