Ural 1297 Palindrome（Manacher或者后缀数组+RMQ-ST）

Posted 2020-10-06 Blackops

 

1297. Palindrome

Time limit: 1.0 second
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.
In addition, it is reasonable to assume that the agent will be sending a very long message, so John has simply to find the longest message satisfying the mentioned property.

Your task is to help John Pupkin by writing a program to find the secret message in the text of a given article. As NPRx8086 ignores white spaces and punctuation marks, John will remove them from the text before feeding it into the program.

Input

The input consists of a single line, which contains a string of Latin alphabet letters (no other characters will appear in the string). String length will not exceed 1000 characters.

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模版题，但是学习到如何输出对应的回文串，即开始坐标=(id-p[id]+1)>>1

代码：

#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;
}

　

玩了下后缀数组，调了半天终于调出来了，注意一些区间合法判断就好了

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 = 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)//枚举以i为中心的奇数情况
        {
            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为靠右位置的偶数情况
            {
                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;
}