URAL 1297 Palindrome (后缀数组+RMQ)
Posted dwtfukgv
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了URAL 1297 Palindrome (后缀数组+RMQ)相关的知识,希望对你有一定的参考价值。
题意:给定一个字符串,求一个最长的回回文子串,多解输出第一个。
析:把字符串翻转然后放到后面去,中间用另一个字符隔开,然后枚举每一个回文串的的位置,对第 i 个位置,那么对应着第二个串的最长公共前缀,
求最长公共子串,可以用RMQ解决。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #define debug() puts("++++"); #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 2000 + 10; const int mod = 1000007; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } struct Array{ int s[maxn], sa[maxn], t[maxn], t2[maxn]; int h[maxn], r[maxn], c[maxn]; int n; int dp[maxn][20]; void init(){ n = 0; memset(sa, 0, sizeof sa); } void build_sa(int m){ int *x = t, *y = t2; for(int i = 0; i < m; ++i) c[i] = 0; for(int i = 0; i < n; ++i) ++c[x[i] = s[i]]; for(int i = 1; i < m; ++i) c[i] += c[i-1]; for(int i = n-1; i >= 0; --i) sa[--c[x[i]]] = i; for(int k = 1; k <= n; k <<= 1){ int p = 0; for(int i = n-k; i < n; ++i) y[p++] = i; for(int i = 0; i < n; ++i) if(sa[i] >= k) y[p++] = sa[i] - k; for(int i = 0; i < m; ++i) c[i] = 0; for(int i = 0; i < n; ++i) ++c[x[y[i]]]; for(int i = 1; i < m; ++i) c[i] += c[i-1]; for(int i = n-1; i >= 0; --i) sa[--c[x[y[i]]]] = y[i]; swap(x, y); p = 1; x[sa[0]] = 0; for(int i = 1; i < n; ++i) x[sa[i]] = y[sa[i-1]] == y[sa[i]] && y[sa[i-1]+k] == y[sa[i]+k] ? p-1 : p++; if(p >= n) break; m = p; } } void getHight(){ int k = 0; for(int i = 0; i < n; ++i) r[sa[i]] = i; for(int i = 0; i < n; ++i){ if(k) --k; int j = sa[r[i]-1]; while(s[i+k] == s[j+k]) ++k; h[r[i]] = k; } } void rmq_init(){ for(int i = 1; i <= n; ++i) dp[i][0] = h[i]; for(int j = 1; (1<<j) <= n; ++j) for(int i = 1; i + (1<<j) <= n; ++i) dp[i][j] = min(dp[i][j-1], dp[i+(1<<j-1)][j-1]); } int query(int L, int R){ L = r[L], R = r[R]; if(L > R) swap(L, R); ++L; int k = int(log(R-L+1) / log(2.0)); return min(dp[L][k], dp[R-(1<<k)+1][k]); } }; char s[maxn]; Array arr; int main(){ while(scanf("%s", s) == 1){ n = strlen(s); arr.init(); for(int i = 0; i < n; ++i) arr.s[arr.n++] = s[i]; arr.s[arr.n++] = 128; for(int i = n-1; i >= 0; --i) arr.s[arr.n++] = s[i]; arr.s[arr.n++] = 0; arr.build_sa(130); arr.getHight(); arr.rmq_init(); int ans = 0, idx; for(int i = 0; i < n; ++i){ int res = max(arr.query(i, 2*n-i) * 2 - 1, arr.query(i, 2*n-i+1) * 2); // odd if(res > ans){ ans = res; idx = i; } } if(ans & 1) for(int i = idx-ans/2; i <= ans/2+idx; ++i) printf("%c", s[i]); else for(int i = idx-ans/2; i < idx+ans/2; ++i) printf("%c", s[i]); printf("\n"); } return 0; }
以上是关于URAL 1297 Palindrome (后缀数组+RMQ)的主要内容,如果未能解决你的问题,请参考以下文章
URAL 1297 Palindrome(后缀数组+ST表)
Ural 1297 Palindrome(Manacher或者后缀数组+RMQ-ST)
URAL 1297 Palindrome (后缀数组+RMQ)