后缀数组小模板 POJ 2774

最新推荐文章于 2020-01-09 10:28:08 发布

THE___BEST

最新推荐文章于 2020-01-09 10:28:08 发布

阅读量423

点赞数

CC 4.0 BY-SA版权

分类专栏：字符串文章标签： poj

本文链接：https://blog.youkuaiyun.com/THE___BEST/article/details/51571361

字符串专栏收录该内容

4 篇文章

订阅专栏

本文介绍了一个模板题的解决方案，该题目涉及字符串处理，通过将两个字符串连接并构建后缀数组来解决。文章提供了完整的C++代码实现，并使用了height数组来辅助计算最长公共前缀。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

模板题，把俩串连起来求后缀数组，通过height数组解当模板用咯

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#define Min(a, b) ((a) < (b)) ? (a) : (b)
#define Max(a, b) ((a) > (b)) ? (a) : (b)
using namespace std;

const int maxn = 200000 + 7;

int cmp(int *r, int a, int b, int l) {
    return (r[a] == r[b]) && (r[a + l] == r[b + l]);
}

int wa[maxn], wb[maxn], WS[maxn], wv[maxn];
int Rank[maxn], height[maxn];

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 *= 2, 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 = 1; 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++;
    }
}

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++);
}

int sa[maxn];
int r[maxn];

char s[maxn], s1[maxn];

int main() {
    scanf("%s", s);
    scanf("%s", s1);
    int len1 = strlen(s), len2 = strlen(s1);
    for (int i = 0; i < len2; ++i) s[len1 + i + 1] = s1[i];
    s[len1] = 128;
    int len = len1 + len2 + 1;
    int n = len + 1;
    for (int i = 0; i < len; ++i) r[i] = s[i];
    r[len1] = 128;
    r[len] = 128;
    DA(r, sa, n + 1, 265);
    calheight(r, sa, n);
    int ans = 0;
    for (int i = 1; i < n; ++i) {
        if (ans < height[i]) {
            int x = Min(sa[i], sa[i - 1]);
            int y = Max(sa[i], sa[i - 1]);
            if (x < len1 && y > len1) ans = height[i];
        }
    }
    printf("%d\n", ans);
    return 0;
}