string string string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 608 Accepted Submission(s): 167
Uncle Mao is a wonderful ACMER. One day he met an easy problem, but Uncle Mao was so lazy that he left the problem to you. I hope you can give him a solution.
Given a string s, we define a substring that happens exactly k times as an important string, and you need to find out how many substrings which are important strings.
The first line contains an integer T (T≤100) implying the number of test cases.
For each test case, there are two lines:
the first line contains an integer k (k≥1) which is described above;
the second line contain a string s (length(s)≤105).
It's guaranteed that ∑length(s)≤2∗106.
∑length(s)≤2∗106.
题目链接：HDU 6194
比赛的时候过的人挺多，可惜刚学$SA$才几天，对$height$的理解还不够深刻，没写出来，只知道大概就是用后缀数组或者更加陌生的自动机什么做吧……
题解也是看别人的：传送门
大致做法就是求刚好出现$k$次的不好求，但是可以转换成至少出现$k$次减去至少出现$k+1$次来算，显然对于至少出现$k$次的问题就可以用$height$数组与RMQ来求，对长度为$k-1$的区间即$(i+1,i+k-1)$进行最小值查询，设最小值为$Min$，那么它对答案的贡献是$Min$，但是可能会多算，因此要减去出现$k+1$的次数，也就是把区间往左移动一个单位$(i,i+k-1)$，往右移动一个单位$(i,i+k)$，减去这两个的最小值$Min1$与$Min2$，此时又会多减掉一些贡献，这些多减掉的贡献就是长度为k的区间即$(i,i+k)$最小值Min3，要将其加回去。代码中用的是LCP函数，直接查询排名为$(l,r)$的区间，可以减少一些边界运算
代码：
#include <stdio.h>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <string>
#include <stack>
#include <cmath>
#include <queue>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
#define fin(name) freopen(name,"r",stdin)
#define fout(name) freopen(name,"w",stdout)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
typedef pair<int, int> pii;
typedef long long LL;
const double PI = acos(-1.0);
const int N = 100010;
int wa[N], wb[N], sa[N], cnt[N];
char s[N];
int height[N], ran[N], dp[N][18];
int len;
inline int cmp(int r[], int a, int b, int d)
{
return r[a] == r[b] && r[a + d] == r[b + d];
}
void DA(int n, int m)
{
int i;
int *x = wa, *y = wb;
for (i = 0; i < m; ++i)
cnt[i] = 0;
for (i = 0; i < n; ++i)
++cnt[x[i] = s[i]];
for (i = 1; i < m; ++i)
cnt[i] += cnt[i - 1];
for (i = n - 1; i >= 0; --i)
sa[--cnt[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)
cnt[i] = 0;
for (i = 0; i < n; ++i)
++cnt[x[y[i]]];
for (i = 1; i < m; ++i)
cnt[i] += cnt[i - 1];
for (i = n - 1; i >= 0; --i)
sa[--cnt[x[y[i]]]] = y[i];
swap(x, y);
x[sa[0]] = 0;
p = 1;
for (i = 1; i < n; ++i)
x[sa[i]] = cmp(y, sa[i - 1], sa[i], k) ? p - 1 : p++;
m = p;
if (p >= n)
break;
}
}
void gethgt(int n)
{
int i, k = 0;
for (i = 1; i <= n; ++i)
ran[sa[i]] = i;
for (i = 0; i < n; ++i)
{
if (k)
--k;
int j = sa[ran[i] - 1];
while (s[j + k] == s[i + k])
++k;
height[ran[i]] = k;
}
}
namespace RMQ
{
void init(int l, int r)
{
int i, j;
for (i = l; i <= r; ++i)
dp[i][0] = height[i];
for (j = 1; l + (1 << j) - 1 <= r; ++j)
for (i = l; i + (1 << j) - 1 <= r; ++i)
dp[i][j] = min(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
}
int query(int l, int r)
{
int len = r - l + 1;
int k = 0;
while (1 << (k + 1) <= len)
++k;
return min(dp[l][k], dp[r - (1 << k) + 1][k]);
}
LL LCP(int rankl, int rankr)
{
if(rankl > rankr)
swap(rankl, rankr);
if (rankl == rankr)
return (LL)len - sa[rankl];
else
return (LL)query(rankl + 1, rankr);
}
}
int main(void)
{
int T, k, i;
scanf("%d", &T);
while (T--)
{
scanf("%d", &k);
scanf("%s", s);
len = strlen(s);
DA(len + 1, 'z' + 1);
gethgt(len);
RMQ::init(1, len);
LL ans = 0;
for (i = 1; i + k - 1 <= len; ++i)
{
ans += RMQ::LCP(i, i + k - 1);
if (i - 1 >= 1)
ans -= RMQ::LCP(i - 1, i + k - 1);
if (i + k <= len)
ans -= RMQ::LCP(i, i + k);
if (i - 1 >= 1 && i + k <= len)
ans += RMQ::LCP(i - 1, i + k);
}
printf("%I64d
", ans);
}
return 0;
}