Boring counting
解题思路
后缀数组。枚举每种长度，对于每个字符串，记录其最大起始位置和最小起始位置，比较是否重合。
代码如下
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N = 1005;
char s[N];
int sa[N], x[N], y[N], c[N];
int n, m;
void get_sa()
{
for(int i = 1; i <= m; i ++) c[i] = 0;
for(int i = 1; i <= n; i ++) c[x[i] = s[i]] ++;
for(int i = 1; i <= m; i ++) c[i] += c[i - 1];
for(int i = n; i >= 1; i --) sa[c[x[i]] --] = i;
for(int k = 1; k <= n; k <<= 1){
int num = 0;
for(int i = n - k + 1; i <= n; i ++) y[++num] = i;
for(int i = 1; i <= n; i ++) if(sa[i] > k) y[++num] = sa[i] - k;
for(int i = 1; i <= m; i ++) c[i] = 0;
for(int i = 1; i <= n; i ++) c[x[i]] ++;
for(int i = 1; i <= m; i ++) c[i] += c[i - 1];
for(int i = n; i >= 1; i --) sa[c[x[y[i]]] --] = y[i], y[i] = 0;
swap(x, y);
num = 1;
x[sa[1]] = num;
for(int i = 2; i <= n; i ++){
if(sa[i] + k <= n && sa[i - 1] + k <= n)
x[sa[i]] = (y[sa[i]] == y[sa[i - 1]] && y[sa[i] + k] == y[sa[i - 1] + k])? num: ++num;
else
x[sa[i]] = ++num;
}
if(num == n)
break;
m = num;
}
}
int height[N], rk[N];
void get_h()
{
int k = 0;
for(int i = 1; i <= n; i ++) rk[sa[i]] = i;
for(int i = 1; i <= n; i ++){
if(rk[i] == 1) continue;
if(k) --k;
int j = sa[rk[i] - 1];
while(i + k <= n && j + k <= n && s[i + k] == s[j + k]) ++k;
height[rk[i]] = k;
}
}
int main()
{
while(scanf("%s", s + 1) != EOF && s[1] != '#'){
n = strlen(s + 1);
m = 'z';
get_sa();
get_h();
ll ans = 0;
for(int k = 1; k <= n / 2; k ++){
int maxx, minn;
maxx = sa[1], minn = sa[1];
for(int i = 2; i <= n; i ++){
if(height[i] < k){
if(maxx - minn >= k) ++ans;
maxx = sa[i], minn = sa[i];
}
else {
maxx = max(maxx, sa[i]);
minn = min(minn, sa[i]);
}
}
if(maxx - minn >= k) ++ ans;
}
printf("%lld\n", ans);
}
return 0;
}