hihocoder-1407 后缀数组二·重复旋律2 不重合 最少重复K次

后缀数组不能直接通过Height得出不重合的公共串。我们可以二分k值,这样连续的Height只要都大于等于k,那他们互相间的k值都大于等于k。每个这样的连续区间查找SA的最大最小值,做差判断是否重合(考虑common prefix=k)。

#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <cstdio>
#include <algorithm>
#include <iomanip>
#include <stdlib.h>
#include <time.h>
#define LL long long
using namespace std;
const LL mod = ;
const LL N = ;
class SF
{
//N:数组大小
public:
int x[N], y[N], c[N];
int Height[N],str[N], SA[N], Rank[N];//Height数组从2开始
int slen;
int m=;//字符集处理大小(传入如果不是数字,需要做位移转换)
bool cmp(int* r, int a, int b, int l) {
return r[a] == r[b] && r[a + l] == r[b + l];
} void Suffix(int n) {
++n;
int i, j, p;
for (i = ; i < m; ++i) c[i] = ;
for (i = ; i < n; ++i) c[x[i] = str[i]]++;
for (i = ; i < m; ++i) c[i] += c[i - ];
for (i = n - ; i >= ; --i) SA[--c[x[i]]] = i;
for (j = ; j <= n; j <<= ) {
p = ;
for (i = n - j; i < n; ++i) y[p++] = i;
for (i = ; i < n; ++i) if (SA[i] >= j) y[p++] = SA[i] - j;
for (i = ; i < m; ++i) c[i] = ;
for (i = ; i < n; ++i) c[x[y[i]]]++; for (i = ; i < m; ++i) c[i] += c[i - ];
for (i = n - ; i >= ; --i) SA[--c[x[y[i]]]] = y[i]; swap(x, y);
p = ; x[SA[]] = ;
for (i = ; i < n; ++i) {
x[SA[i]] = cmp(y, SA[i - ], SA[i], j) ? p - : p++;
}
if (p >= n)break;
m = p;
} int k = ;
n--;
for (i = ; i <= n; ++i) Rank[SA[i]] = i;
for (i = ; i < n; ++i) {
if (k)--k;
j = SA[Rank[i] - ];
while (str[i + k] == str[j + k])++k;
Height[Rank[i]] = k;
//cout << k << endl;
}
}
void init(vector<int> &vv)
{
slen = vv.size();
for (int i = ; i < slen; i++)
str[i] = vv[i]+;//如果是字符,映射成从1开始的序列
str[slen] = ;//0作为结束符,防止越界
Suffix(slen);
}
struct nod
{
int mx, mi;
};
int ans;
bool ok(int k)
{
int temp = ;
int mx = , mi = ;
bool f = false;
for (int i = ; i <= slen; i++)
{
if (Height[i] >= k)
{
mx = max(mx, max(SA[i - ], SA[i]));
mi = min(mi, min(SA[i - ], SA[i]));
}
else
{
mx = ;
mi = ;
}
if (mx - mi >= k)f = true;
}
return f;
}
int go(int l, int r)
{
for (; r >= l; r--)
if (ok(r))return r;
return ;
}
int bins(int l, int r)
{
while (l <= r)
{
if(r-l<=)
return go(l, r);
int mid = (l + r) / ;
if (ok(mid))
l = mid;
else r = mid - ;
}
}
LL solve()
{
ans=bins(, slen);
return ans;
}
}sf;
LL dp[][];
LL n;
int main() {
cin.sync_with_stdio(false);
while (cin >> n)
{
vector<int> v;
for (int i = ; i < n; i++)
{
int num;
cin >> num;
v.push_back(num);
}
sf.init(v);
cout << sf.solve() << endl;
}
return ;
}
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