求出长度为1， 2， 3， 4， 5....的字符串最大出现次数 后缀数组 UVA 11855 - Buzzwords

题目链接

题意：给定一个文本，求出长度为1， 2， 3， 4， 5....的字符串最大出现次数，一直找到出现次数不大于1为止

思路：后缀数组，把空格去掉得到一个字符串，构造height数组，然后就是后缀数组的简单应用了，找连续一段最长的段并且都大于当前正在查找的长度

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>
using namespace std;

const int MAXLEN = 1005;

struct Suffix {

int s[MAXLEN];
int sa[MAXLEN], t[MAXLEN], t2[MAXLEN], c[MAXLEN], n;
int rank[MAXLEN], height[MAXLEN];

void build_sa(int m) {
n++;
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];
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;
}
n--;
}

void getHeight() {
int i, j, k = 0;
for (i = 1; 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;
}
}

void init(string str) {
n = 0;
for (int i = 0; i < str.length(); i++) if (str[i] != ' ')
s[n++] = str[i] - 'A' + 1;
s
= '\0';
}

void solve() {
build_sa(27);
getHeight();
int len = 1;
while (1) {
int ans = 1;
for (int i = 1; i <= n; i++) {
if (n - sa[i] < len) continue;
int cnt = 1;
while (i < n && height[i + 1] >= len) {
i++;
cnt++;
}
ans = max(ans, cnt);
}
if (ans == 1) break;
printf("%d\n", ans);
len++;
}
printf("\n");
}
} gao;

string str;

int main() {
while (getline(cin, str) && str != "") {
gao.init(str);
gao.solve();
}
return 0;
}