Ural 1297 Palindrome（Manacher或者後綴數組+RMQ-ST）
阿新 • • 發佈：2017-09-12
奇數 receive scribe exce rmq channel ignore uri turn
Memory limit: 64 MB
The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots» security service would have already started an undercover operation to establish the agent’s identity, but, fortunately, the letter describes communication channel the agent uses. He will publish articles containing stolen data to the “Solaris” almanac. Obviously, he will obfuscate the data, so “Robots Unlimited” will have to use a special descrambler (“Robots Unlimited” part number NPRx8086, specifications are kept secret). Having read the letter, the “U.S. Robots” president recalled having hired the “Robots Unlimited” ex-employee John Pupkin. President knows he can trust John, because John is still angry at being mistreated by “Robots Unlimited”. Unfortunately, he was fired just before his team has finished work on the NPRx8086 design. So, the president has assigned the task of agent’s message interception to John. At first, John felt rather embarrassed, because revealing the hidden message isn’t any easier than finding a needle in a haystack. However, after he struggled the problem for a while, he remembered that the design of NPRx8086 was still incomplete. “Robots Unlimited” fired John when he was working on a specific module, the text direction detector. Nobody else could finish that module, so the descrambler will choose the text scanning direction at random. To ensure the correct descrambling of the message by NPRx8086, agent must encode the information in such a way that the resulting secret message reads the same both forwards and backwards.
1297. Palindrome
Time limit: 1.0 secondMemory limit: 64 MB
The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unlimited» has infiltrated into “U.S. Robotics”. «U.S. Robots» security service would have already started an undercover operation to establish the agent’s identity, but, fortunately, the letter describes communication channel the agent uses. He will publish articles containing stolen data to the “Solaris” almanac. Obviously, he will obfuscate the data, so “Robots Unlimited” will have to use a special descrambler (“Robots Unlimited” part number NPRx8086, specifications are kept secret). Having read the letter, the “U.S. Robots” president recalled having hired the “Robots Unlimited” ex-employee John Pupkin. President knows he can trust John, because John is still angry at being mistreated by “Robots Unlimited”. Unfortunately, he was fired just before his team has finished work on the NPRx8086 design. So, the president has assigned the task of agent’s message interception to John. At first, John felt rather embarrassed, because revealing the hidden message isn’t any easier than finding a needle in a haystack. However, after he struggled the problem for a while, he remembered that the design of NPRx8086 was still incomplete. “Robots Unlimited” fired John when he was working on a specific module, the text direction detector. Nobody else could finish that module, so the descrambler will choose the text scanning direction at random. To ensure the correct descrambling of the message by NPRx8086, agent must encode the information in such a way that the resulting secret message reads the same both forwards and backwards.
Input
Output
The longest substring with mentioned property. If there are several such strings you should output the first of them.Sample
| input | output |
|---|---|
ThesampletextthatcouldbereadedthesameinbothordersArozaupalanalapuazorA |
ArozaupalanalapuazorA |
Manacher模版題，但是學習到如何輸出對應的回文串，即開始坐標=(id-p[id]+1)>>1
代碼：
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define INF 0x3f3f3f3f
#define MM(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long LL;
const int N = 1010;
char s[N], ss[2 * N];
int p[2 * N];
inline int manacher(char s[])
{
int mx = p[0] = 0, idd, len = strlen(s), S = 0, L = 0;
for (int i = 1; i < len; i++)
{
p[i] = 1;
if (mx > i)
{
p[i] = p[2 * idd - i];
if (mx - i < p[i])
p[i] = mx - i;
}
while (s[i - p[i]] == s[i + p[i]])
p[i]++;
if (i + p[i] > mx)
{
mx = i + p[i];
idd = i;
if (p[i] > S)
S = p[i];
}
}
return S;
}
int main(void)
{
int i, j;
while (~scanf("%s", s))
{
ss[0] = ‘$‘;
int len = strlen(s);
for (i = 0; i < len; i++)
{
ss[2 * i + 1] = ‘#‘;
ss[2 * i + 2] = s[i];
}
ss[2 * len + 1] = ‘#‘;
int LEN = manacher(ss) - 1;
int idd = 0;
for (i = 1; i < 2 * len + 1; i++)
{
if (p[i] > p[idd])
idd = i;
}
int cnt = 0;
for (i = (idd - p[idd] + 1) >> 1; cnt < LEN; i++, cnt++)
putchar(s[i]);
putchar(‘\n‘);
MM(s);
MM(ss);
MM(p);
}
return 0;
}
玩了下後綴數組，調了半天終於調出來了，註意一些區間合法判斷就好了
代碼：
#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 = 220010;
int wa[N], wb[N], cnt[N], sa[N];
int ran[N], height[N];
char s[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][18];
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)
{
if (l > r)
swap(l, r);
if (l == r)
return len - sa[l];
return ask(l + 1, r);
}
}
int main(void)
{
int i;
while (~scanf("%s", s))
{
int len = strlen(s);
for (i = 0; i < len; ++i)
s[len + i] = s[len - 1 - i];
s[len << 1] = ‘\0‘;
DA(len << 1 | 1, ‘z‘ + 1);
gethgt(len << 1);
SG::init(1, len << 1);
int ans = 1;
int pos = 0;
for (i = 0; i < len; ++i)//枚舉以i為中心的奇數情況
{
if (i + 1 < len && len * 2 - 1 - (i - 1) < len * 2)
{
int lcp = SG::LCP(ran[i + 1], ran[len * 2 - 1 - (i - 1)], len);
lcp = min({lcp, i, len - 1 - i});
int Ans = lcp * 2 + 1;
if (Ans > ans || (Ans == ans && i - lcp < pos))
{
ans = Ans;
pos = i - lcp;
}
}
if (len * 2 - 1 - (i - 1) < len * 2)//以i為靠右位置的偶數情況
{
int lcp = SG::LCP(ran[i], ran[len * 2 - 1 - (i - 1)], len);
lcp = min({lcp, len - i, i});
int Ans = lcp * 2;
if (Ans > ans || (Ans == ans && i - lcp < pos))
{
ans = Ans;
pos = i - lcp;
}
}
}
printf("%d\n", ans);
}
return 0;
}
Ural 1297 Palindrome（Manacher或者後綴數組+RMQ-ST）