标签：its 字符 inpu 题目 gre height turn ext cst
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10461 | Accepted: 3234 |
Description
The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "ababa" is 1.
Given a string containing lowercase letters, you are to find a substring of it with maximum repetition number.
Input
The input consists of multiple test cases. Each test case contains exactly one line, which
gives a non-empty string consisting of lowercase letters. The length of the string will not be greater than 100,000.
The last test case is followed by a line containing a ‘#‘.
Output
For each test case, print a line containing the test case number( beginning with 1) followed by the substring of maximum repetition number. If there are multiple substrings of maximum repetition number, print the lexicographically smallest one.
Sample Input
ccabababc daabbccaa #
Sample Output
Case 1: ababab Case 2: aa
题目链接：POJ 3693
把所有可能构成的最多重复次数的子串所对应的循环节大小存下来，然后枚举$SA[i]$与循环节大小，如果刚好枚举到了这个$SA[i]$对应的就是这个循环节大小，那么就是最小的字典序解了，因为$SA[]$是按照字典序排的，当然要注意单个字符这种字符串，因此一开始要把长度定为1。
另外这题数据非常弱，实际上枚举的时候需要加一些边界防止越界问题。可以试一下这些数据：
kabhvlkba
slgnaebbga
lajnbabab
kabkbakbvkab
akbakabka
akjbakjbajkba
akjbakbaiajklbna
kljdfnbisn
akbvkab
答案应该是
a
b
abab
a
a
akjbakjb
a
b
a
代码：
#include <stdio.h>
#include <iostream>
#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], cnt[N], sa[N];
int ran[N], height[N];
char s[N];
int pos[N];
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 (m >= 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 SG
{
int dp[N][17];
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 ask(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]);
}
int LCP(int l, int r, int len)
{
l = ran[l], r = ran[r];
if (l > r)
swap(l, r);
if (l == r)
return len - sa[l];
return ask(l + 1, r);
}
}
int main(void)
{
int T = 0, len, i, j;
while (~scanf("%s", s) && s[0] != ‘#‘)
{
len = strlen(s);
DA(len + 1, 130);
gethgt(len);
SG::init(1, len);
int ans = 1;
int sz = 0;
for (int L = 1; L < len; ++L)
{
for (i = 0; i + L < len; i += L)
{
int lcp = SG::LCP(i, i + L, len);
int cnt = lcp / L + 1;
int j = i - (L - lcp % L);
if (j >= 0 && lcp % L != 0 && SG::LCP(j , j + L, len) / L + 1 > cnt)
++cnt;
if (cnt > ans)
{
ans = cnt;
sz = 0;
pos[sz++] = L;
}
else if (cnt == ans)
pos[sz++] = L;
}
}
int length = 1;
int st = 0;
int flag = 0;
for (i = 1; i <= len && !flag; ++i)
{
for (j = 0; j < sz; ++j)
{
int unit = pos[j];
if (sa[i] + unit < len && SG::LCP(sa[i], sa[i] + unit, len) >= (ans - 1) * unit)
{
length = ans * unit;
st = sa[i];
flag = 1;
break;
}
}
}
printf("Case %d: ", ++T);
for (i = st; i < st + length; ++i)
putchar(s[i]);
puts("");
}
return 0;
}
POJ 3693 Maximum repetition substring（最多重复次数的子串）
标签：its 字符 inpu 题目 gre height turn ext cst
原文地址：http://www.cnblogs.com/Blackops/p/7520818.html
