poj 2774(后缀数组)

题意：每次给出两个串，问两个串的最长公共子串的长度。

题解：后缀数组算法合集之《后缀数组——处理字符串的有力工具》这里讲解的很好，把两个串连接起来，然后求这个串的后缀数组，和对应的height[i]的最大值，但主要要满足sa[i]和sa[i-1]在不同的串中。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 200005;
int wa
, wb
, ws
, wv
, sa
;
int rank
, height
, s
;
char s1
, s2
;

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;
//计数排序
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) {  //j是长度
//求第二关键字(可以借助之前计算过的sa)
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;
//新一轮排序
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];
//求第一关键字
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 main() {
while (scanf("%s%s", s1, s2) == 2) {
int len1 = strlen(s1), len2 = strlen(s2), cnt = 0;
for (int i = 0; i < len1; i++)
s[cnt++] = s1[i] - 'a' + 1;
s[cnt++] = 27;
for (int i = 0; i < len2; i++)
s[cnt++] = s2[i] - 'a' + 1;
s[cnt] = 0;
DA(s, sa, cnt + 1, 30);
calheight(s, sa, cnt);
int res = 0;
for (int i = 2; i < cnt; i++)
if (height[i] > res) {
if (sa[i - 1] >= 0 && sa[i - 1] < len1 && sa[i] > len1)
res = height[i];
else if (sa[i] >= 0 && sa[i] < len1 && sa[i - 1] > len1)
res = height[i];
}
printf("%d\n", res);
}
return 0;
}