| 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;
}
Ural 1297 Palindrome（Manacher或者后缀数组+RMQ-ST）
原文：http://www.cnblogs.com/Blackops/p/5766356.html