URAL

1989. Subpalindromes

Time limit: 0.5 second
Memory limit: 64 MB

You have a string and queries of two types:

1. replace i’th character of the string by character a;
2. check if substring sj...sk is a palindrome.

Input

The first line contains a string consisting of n small English letters. The second line contains an integer mthat is the number of queries  (5 ≤ n, m ≤ 105). The next m lines contain the queries. 
Each query has either form “change i a”, or “palindrome? j k”, where i, j, k are integers (1 ≤ i ≤ n; 1 ≤ j ≤ k ≤ n), and character a is a small English letter.

Output

To all second type queries, you should output “Yes” on a single line if substring sj...sk is a palindrome and “No” otherwise.

Sample

inputoutput

abcda5palindrome? 1 5palindrome? 1 1change 4 bpalindrome? 1 5palindrome? 2 4

NoYesYesYes

Problem Author: Mikhail Rubinchik
Problem Source: Open Ural FU Championship 2013

Tags: palindromes  (

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Difficulty: 310    Printable version    Submit solution    Discussion (5)
All submissions (5091)    All accepted submissions (1022)    Solutions rating (432)

思路：

比较询问区间正向hash 值和反向hash值，相等则说明是回文字符串。

对于求区间的hash值，因为时限只有500ms，所以至少要在logn的复杂度求出hash值，对此线段树和数组数组都可以

线段树版：

#include <iostream>#include <vector>#include <string>#include <algorithm>#include <cstdio>#include <cstring>#include <ctime>#include <cstdlib>#include <set>#include <map>#include <cctype>#include <list>#include <cmath>#include <bitset>#include <queue>#include <stack>#include <sstream>#include <functional>#include <cassert>using namespace std;#define tmax(a, b) if (b > a) a = b#define tmin(a, b) if (b < a) a = b#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1#define par(a) a >> 1char inc() { char _[10]; scanf("%s", _); return _[0]; }int ini() { int _; scanf("%d", &_); return _; }long long inll() { long long _; scanf("%I64d", &_); return _; }double ind() { double _; scanf("%lf", &_); return _; }string ins() { string _; cin >> _; return _; }int inl(char _[]) { if (!fgets(_, (int)1e8, stdin)) return -1; int i = strlen(_); if (_[i - 1] == '\n') _[--i] = 0; return i; }typedef pair<int, int> pii;typedef pair<char, char> pcc;typedef long long LL;typedef unsigned long long ULL;const int inf = 0x3f3f3f3f;const LL lnf = 0x3f3f3f3f3f3f3f3f;const double pi = 3.14159265358979323846;const double eps = 1e-8;const int mod = 100007;const int maxn = 1e5 + 10;char str[maxn];int len;ULL d[maxn];ULL sum[2][maxn << 2];void PushUp(int rt, int flag) {sum[flag][rt] = sum[flag][rt << 1] + sum[flag][rt << 1 | 1];}void Build(int l, int r, int rt, int flag) {if (l == r) {sum[flag][rt] = d[l] * (str[l] - 'a');return;}int m = (l + r) >> 1;Build(lson, flag);Build(rson, flag);PushUp(rt, flag);}void Update(int p, ULL v, int l, int r, int rt, int flag) {if (l == r) {sum[flag][rt] = v;return;}int m = (l + r) >> 1;if (p <= m) Update(p, v, lson, flag);else Update(p, v, rson, flag);PushUp(rt, flag);}ULL Query(int L, int R, int l, int r, int rt, int flag) {if (L <= l && r <= R) {return sum[flag][rt];}int m = (l + r) >> 1;ULL res = 0;if (L <= m) res += Query(L, R, lson, flag);if (m < R) res += Query(L, R, rson, flag);return res;}int main() {int CAS = 0;//std::ios::sync_with_stdio(0);//std::cin.tie(0);#ifdef NIGHT_13//freopen("myout.txt", "w", stdout);freopen("in.txt", "r", stdin);int time_night_13 = clock();#endif // NIGHT_13scanf("%s", str + 1);len = strlen(str + 1);d[1] = 1;for (int i = 2; i <= len; ++i) d[i] = d[i - 1] * 29;Build(1, len, 1, 0);reverse(str + 1, str + 1 + len);Build(1, len, 1, 1);int T = ini();while (T--) {char op[100];scanf("%s", op);if (op[0] == 'p') {int l, r;scanf("%d%d", &l, &r);ULL ls = Query(l, r, 1, len, 1, 0);ULL rs = Query(len + 1 - r, len + 1 - l, 1, len, 1, 1);int k1 = l - 1, k2 = len - r;if (k1 < k2) ls *= d[k2 - k1 + 1];if (k1 > k2) rs *= d[k1 - k2 + 1];puts(ls == rs ? "Yes" : "No");} else {int a;scanf("%d%s", &a, op);Update(a, d[a] * (op[0] - 'a'), 1, len, 1, 0);Update(len + 1 - a, d[len + 1 - a] * (op[0] - 'a'), 1, len, 1, 1);}}#ifdef NIGHT_13fprintf(stderr, "\nTime: %dms\n", (int)clock() - time_night_13);#endif // NIGHT_13return 0;}

树状数组版：

#include <iostream>#include <vector>#include <string>#include <algorithm>#include <cstdio>#include <cstring>#include <ctime>#include <cstdlib>#include <set>#include <map>#include <cctype>#include <list>#include <cmath>#include <bitset>#include <queue>#include <stack>#include <sstream>#include <functional>#include <cassert>using namespace std;#define tmax(a, b) if (b > a) a = b#define tmin(a, b) if (b < a) a = b#define lson l, m, rt << 1#define rson m + 1, r, rt << 1 | 1#define par(a) a >> 1char inc() { char _[10]; scanf("%s", _); return _[0]; }int ini() { int _; scanf("%d", &_); return _; }long long inll() { long long _; scanf("%I64d", &_); return _; }double ind() { double _; scanf("%lf", &_); return _; }string ins() { string _; cin >> _; return _; }int inl(char _[]) { if (!fgets(_, (int)1e8, stdin)) return -1; int i = strlen(_); if (_[i - 1] == '\n') _[--i] = 0; return i; }typedef pair<int, int> pii;typedef pair<char, char> pcc;typedef long long LL;typedef unsigned long long ULL;const int inf = 0x3f3f3f3f;const LL lnf = 0x3f3f3f3f3f3f3f3f;const double pi = 3.14159265358979323846;const double eps = 1e-8;const int mod = 100007;const int maxn = 1e5 + 10;char str[maxn];int len;ULL d[maxn] = {0, 1};ULL bt[maxn][2];void Add(int x, int k, ULL v) {while (x <= len) {bt[x][k] += v;x += -x & x;}}ULL Query(int x, int k) {ULL res = 0;while (x) {res += bt[x][k];x -= -x & x;}return res;}int main() {int CAS = 0;//std::ios::sync_with_stdio(0);//std::cin.tie(0);#ifdef NIGHT_13//freopen("myout.txt", "w", stdout);freopen("in.txt", "r", stdin);int time_night_13 = clock();#endif // NIGHT_13scanf("%s", str + 1);len = strlen(str + 1);for (int i = 2; i <= len; ++i) d[i] = d[i - 1] * 29;for (int i = 1; i <= len; ++i) {Add(i, 0, d[i] * str[i]);Add(len + 1 - i, 1, d[len + 1 - i] * str[i]);}int T = ini();while (T--) {char op[100];scanf("%s", op);if (op[0] == 'p') {int l, r;scanf("%d%d", &l, &r);ULL ls = Query(r, 0) - Query(l - 1, 0);ULL rs = Query(len + 1 - l, 1) - Query(len - r, 1);int k1 = l - 1, k2 = len - r;if (k1 < k2) ls *= d[k2 - k1 + 1];if (k1 > k2) rs *= d[k1 - k2 + 1];puts(ls == rs ? "Yes" : "No");} else {int a;scanf("%d%s", &a, op);Add(a, 0, d[a] * (op[0] - str[a]));Add(len + 1 - a, 1, d[len + 1 - a] * (op[0] - str[a]));str[a] = op[0];}}#ifdef NIGHT_13fprintf(stderr, "\nTime: %dms\n", (int)clock() - time_night_13);#endif // NIGHT_13return 0;}