hdu 1544 统计回文子串的个数

题意：给一个长度不超过5000的字符串，统计回文子串的个数。

思路：枚举回文串的中点和长度（长度还要分奇偶），统计个数即可，复杂度为O(n^2)。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 5001;
char str[N];

int solve()
{
    int ans = 0, len = strlen(str);
    for ( int i = 0; i < len; i++ )
    {
        for ( int j = 0; i - j >= 0 && i + j < len; j++ )
        {
            if ( str[i - j] == str[i + j] )
            {
                ans++;
            }
            else
            {
                break;
            }
        }
        if ( str[i] != str[i + 1] ) continue;
        for ( int j = 0; i - j >= 0 && i + 1 + j < len; j++ )
        {
            if ( str[i - j] == str[i + 1 + j] )
            {
                ans++;
            }
            else
            {
                break;
            }
        }
    }
    return ans;
}

int main ()
{
    while ( scanf("%s", str) != EOF )
    {
        printf("%d
", solve());
    }
    return 0;
}

manacher算法的解法：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 5007;
char str[N];
char tmp[N << 1];
int len[N << 1];

int convert( char * st, char * dst )
{
    int l = strlen(st);
    dst[0] = '@';
    for ( int i = 1; i <= 2 * l; i += 2 )
    {
        dst[i] = '#';
        dst[i + 1] = st[i / 2];
    }
    dst[2 * l + 1] = '#';
    dst[2 * l + 2] = 0;
    return 2 * l + 1;
}

int manacher( char * st, char * dst )
{
    int l = convert( st, dst );
    int mx = 0, ans = 0, po = 0;
    for ( int i = 1; i <= l; i++ )
    {
        if ( mx > i )
        {
            len[i] = min( mx - i, len[2 * po - i] );
        }
        else
        {
            len[i] = 1;
        }
        while ( dst[i - len[i]] == dst[i + len[i]] )
        {
            len[i]++;
        }
        if ( len[i] + i > mx )
        {
            mx = len[i] + i;
            po = i;
        }
        //ans = max( ans, len[i] );
        ans += len[i] / 2;
    }
    //return ans - 1;
    return ans;
}

int main ()
{
    while ( scanf("%s", str) != EOF )
    {
        int o = manacher( str, tmp );
        printf("%d
", o);
    }
    return 0;
}

贴一个过不了此题但自认为写的不错的一个O(n^3)的算法：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

const int N = 5001;
char str[N];

int judge( int l, int r )
{
    while ( l < r )
    {
        if ( str[l++] != str[r--] ) return 0;
    }
    return 1;
}

int solve()
{
    int ans = 0, len = strlen(str);
    for ( int i = 0; i < len; i++ )
    {
        for ( int j = i; j < len; j++ )
        {
            ans += judge( i, j );
        }
    }
    return ans;
}

int main ()
{
    while ( scanf("%s", str) != EOF )
    {
        printf("%d
", solve());
    }
    return 0;
}