标签:class com return pre string 允许 can memset http
传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=1031
保存后缀数组模版。
其实如果数据范围小一点,或者空间限制再大一点,或者甚至只要字母表再小一点就可以使用后缀自动机了,只可惜空间不允许,就用后缀数组!
其实还是不是很理解代码,是否该当成黑盒代码背下来呢?
#include <cstdio>
#include <cstring>
#include <algorithm>
const int maxn = 200005;
int n, mx, sa[maxn], t1[maxn], t2[maxn], c[maxn], *x, *y;
char s[maxn];
int main(void) {
scanf("%s", s);
n = strlen(s);
memcpy(s + n, s, n - 1);
n = (n << 1) - 1;
x = t1;
y = t2;
mx = 128;
memset(c, 0, sizeof c);
for (int i = 0; i < n; ++i) {
++c[x[i] = s[i]];
}
for (int i = 1; i < mx; ++i) {
c[i] += c[i - 1];
}
for (int i = n - 1; ~i; --i) {
sa[--c[x[i]]] = i;
}
int p;
for (int k = 1; k <= n; k <<= 1) {
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;
}
}
memset(c, 0, sizeof c);
for (int i = 0; i < n; ++i) {
++c[x[y[i]]];
}
for (int i = 0; i < mx; ++i) {
c[i] += c[i - 1];
}
for (int i = n - 1; ~i; --i) {
sa[--c[x[y[i]]]] = y[i];
}
std::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;
}
mx = p;
}
int tem = (n + 1) >> 1;
for (int i = 0; i < n; ++i) {
if (sa[i] < tem) {
printf("%c", s[sa[i] + tem - 1]);
}
}
return 0;
}
_bzoj1031 [JSOI2007]字符加密Cipher【后缀数组】
标签:class com return pre string 允许 can memset http
原文地址:http://www.cnblogs.com/ciao-sora/p/6166065.html