URAL 1297 Palindrome 尾碼數組+RMQ

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本題是利用尾碼數組求最長的迴文串。

方法是將字串反轉之後拼接到原來的字串末尾，中間用一個沒有出現過的分割符隔開，原因是防止最長公用首碼橫跨兩個串。

之後分別枚舉迴文串的中點，以及迴文串長度是奇數還是偶數，看一下對應位置的最長公用首碼即可。

這裡的求最長公用首碼要處理RMQ問題，線段樹固然可以解決，但是顯然ST 演算法更加快一些。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 5005;//以下是倍增法求尾碼數組int wa[maxn], wb[maxn], wv[maxn], ws[maxn];int cmp(int *r, int a, int b, int l) { return r[a] == r[b] && r[a + l] == r[b + l]; }void da(int *r, int *sa, int n, int m) {    int i, j, p, *x = wa, *y = wb, *t;    for(i = 0; i < m; i++) ws[i] = 0;    for(i = 0; i < n; i++) ws[x[i] = r[i]]++;    for(i = 1; i < m; i++) ws[i] += ws[i - 1];    for(i = n - 1; i >= 0; i--) sa[--ws[x[i]]] = i;    for(j = 1, p = 1; p < n; j <<= 1, 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++) ws[i] = 0;        for(i = 0; i < n; i++) ws[wv[i]]++;        for(i = 0; i < m; i++) ws[i] += ws[i - 1];        for(i = n - 1; i >= 0; i--) sa[--ws[wv[i]]] = y[i];        for(t = x, x = y, y = t, 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++;    }}//以下是求解height 數組int height[maxn], Rank[maxn];void calheight(int *r, int *sa, 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)         for(k ? k-- : 0, j = sa[Rank[i] - 1];                 r[i + k] == r[j + k]; k++) ;}char buf[maxn];int str[maxn], sa[maxn], len, n;int minv[maxn][20];void init_RMQ() {    for(int i = 0; i <= len; i++) minv[i][0] = height[i];    for(int j = 1; (1 << j) <= len + 1; j++) {        for(int i = 0; i + (1 << j) - 1 <= len; i++) {            minv[i][j] = min(minv[i][j - 1], minv[i + (1 << (j - 1))][j - 1]);        }    }}int query(int ql, int qr) {    if(ql > qr) swap(ql, qr);    ql++;    int k = 0;    while((1 << (k + 1)) <= qr - ql + 1) k++;    return min(minv[ql][k], minv[qr - (1 << k) + 1][k]);}int main() {    while(scanf("%s", buf) != EOF) {        n = len = strlen(buf);        for(int i = 0; i < len; i++) str[i] = buf[i] + 1;        str[len++] = ‘*‘ + 1;        for(int i = n - 1; i >= 0; i--) str[len++] = buf[i] + 1;        str[len] = 0;        da(str, sa, len + 1, 200);        calheight(str, sa, len);        init_RMQ();        int anslen = 1, anspos = 0;        for(int i = 0; i < n; i++) {            int nowval = query(Rank[i], Rank[len - i - 1]);            int nowlen = nowval * 2 - 1;            if(nowlen > anslen) {                anslen = nowlen; anspos = i - nowval + 1;            }            if(i == 0) continue;            nowval = query(Rank[i], Rank[len - i]);            nowlen = nowval * 2;            if(nowlen > anslen) {                anslen = nowlen; anspos = i - nowval;            }        }        for(int i = 0; i < anslen; i++) putchar(buf[i + anspos]);        puts("");    }    return 0;}

 

URAL 1297 Palindrome 尾碼數組+RMQ