后缀数组 POJ 3974 Palindrome && URAL 1297 Palindrome

题目链接

题意：求给定的字符串的最长回文子串

分析：做法是构造一个新的字符串是原字符串+反转后的原字符串(这样方便求两边回文的后缀的最长前缀)，即newS = S + ‘$‘ + revS，枚举回文串中心位置，RMQ询问LCP = min (height[rank[l]+1] to height[rank[r]])，注意的是RMQ传入参数最好是后缀的位置，因为它们在树上的顺序未知，且左边还要+1。

#include <cstdio>
#include <algorithm>
#include <cstring>

const int N = 1e3 + 10;
const int D = 21;
char s[N];
int sa[N<<1], t[N<<1], t2[N<<1], c[N<<1], n;
int rank[N<<1], height[N<<1];
int dp[N<<1][D];

void da(char *s, int m = 256) {
    int i, *x = t, *y = t2;
    for (i=0; i<m; ++i) c[i] = 0;
    for (i=0; i<n; ++i) c[x[i]=s[i]]++;
    for (i=1; i<m; ++i) c[i] += c[i-1];
    for (i=n-1; i>=0; --i) sa[--c[x[i]]] = i;
    for (int k=1; k<=n; k<<=1) {
        int p = 0;
        for (i=n-k; i<n; ++i) y[p++] = i;
        for (i=0; i<n; ++i) if (sa[i] >= k) y[p++] = sa[i] - k;
        for (i=0; i<m; ++i) c[i] = 0;
        for (i=0; i<n; ++i) c[x[y[i]]]++;
        for (i=0; i<m; ++i) c[i] += c[i-1];
        for (i=n-1; i>=0; --i) sa[--c[x[y[i]]]] = y[i];
        std::swap (x, y);
        p = 1; x[sa[0]] = 0;
        for (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;
    }
    //height[i] = LCP (s[sa[i-1]], s[sa[i]]);
    int j, k = 0;
    for (i=0; i<n; ++i) rank[sa[i]] = i;
    for (i=0; i<n; ++i) {
        if (k) k--;
        int j = sa[rank[i]-1];
        while (s[i+k] == s[j+k]) k++;
        height[rank[i]] = k;
    }
}

int query_RMQ(int l, int r) {
    l = rank[l]; r = rank[r];
    if (l > r) {
        std::swap (l, r);
    }
    l++;
    int k = 0; while (1<<(k+1) <= r - l + 1) k++;
    return std::min (dp[l][k], dp[r-(1<<k)+1][k]);
}
void init_RMQ(int *height) {
    //memset (dp, 0, sizeof (dp));
    for (int i=0; i<=n; ++i) {
        dp[i][0] = height[i];
    }
    for (int j=1; (1<<j)<=n; ++j) {
        for (int i=0; i+(1<<j)<=n; ++i) {
            dp[i][j] = std::min (dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
        }
    }
}

int run(char *s) {
    int len = strlen (s);
    n = len;
    s[n++] = ‘$‘;
    for (int i=len-1; i>=0; --i) {
        s[n++] = s[i];
    }
    s[n++] = ‘\0‘;
    da (s);
    init_RMQ (height);
    int ret = 0;
    for (int i=0; i<len; ++i) {
        int l = query_RMQ (i, n-i-2); //奇数
        ret = std::max (ret, 2 * l - 1);
        l = query_RMQ (i, n-i-1);  //偶数
        ret = std::max (ret, 2 * l);
    }
    return ret;
}

int main() {
    int cas = 0;
    while (scanf ("%s", s) == 1) {
        if (strcmp (s, "END") == 0) {
            break;
        }
        printf ("Case %d: %d\n", ++cas, run (s));
    }
    return 0;
}

URAL 1297 题目没什么区别，数据规模小了，还要输出回文串。

核心代码

int best = 0;
from = 0; to = 1;
for (int i=0; i<len; ++i) {
    int l = query_RMQ (i, n-i-1);
    if (best < 2 * l - 1) {
        from = i - l + 1; to = i + l;
        best = 2 * l - 1;
    }
    l = query_RMQ (i, n-i);
    if (best < 2 * l) {
        from = i - l; to = i + l;
        best = 2 * l;
    }
}