HDU 3086 马拉车模板

模板，但是对这个算法还是不太清楚,真实不明觉厉....

#include <iostream>
#include <cstdio>
#include <string.h>
#pragma warning ( disable : 4996 )
using namespace std;

inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
const int inf = 0x3f3f3f3f;
const int maxn = 1e5+1e4+5;

char str[maxn];
char nstr[maxn<<1];
int maxlen[maxn<<1];
int len, plen, ans;

void init()
{
    memset( nstr, 0, sizeof(nstr) );
    memset( maxlen, 0, sizeof(maxlen) );
    len = strlen(str);    
    nstr[0] = '$'; nstr[1] = '#';

    int j = 2;
    for ( int i = 0; i < len; i++ )
    {
        nstr[j++] = str[i];
        nstr[j++] = '#';
    }
    nstr[j] = '';
    plen = j;
}

void solve()
{
    ans = -1;
    int id, mx = 0;

    for ( int i = 1; i < plen; i++ )
    {
        if( i < mx )
            maxlen[i] = Min( maxlen[2*id-i], mx-i );
        else
            maxlen[i] = 1;

        while( nstr[i-maxlen[i]] == nstr[i+maxlen[i]] )
            maxlen[i]++;

        if ( mx < i + maxlen[i] )
        {
            id = i;
            mx = i + maxlen[i];
        }
        ans = Max(ans, maxlen[i]-1);
    }
}

int main()
{
    while ( ~scanf("%s", str) )
    {
        init();
        solve();
        printf( "%d
", ans );
    }
    return 0;
}

 又做了一道几乎模板的题（吉哥系列故事——完美队形II），终于对马拉车有点理解了，这算法实在太巧妙了！

和模板几乎一样，只不过增多了个左半边要升序排列罢了

#include <iostream>
#include <cstdio>
#include <string.h>
#pragma warning ( disable : 4996 )
using namespace std;

inline int Max(int a,int b) { return a>b?a:b; }
inline int Min(int a,int b) { return a>b?b:a; }
const int inf = 0x3f3f3f3f;
const int maxn = 1e5+5;

int maxl[maxn<<1];
int str[maxn], nstr[maxn<<1];
int N, ans, plen;

void init()
{
    memset( maxl, 0, sizeof(maxl) );
    memset( nstr, 0, sizeof(nstr) );
    nstr[0] = 0; nstr[1] = -1;

    int j = 2;
    for( int i = 0; i < N; i++ )
        { nstr[j++] = str[i]; nstr[j++] = -1; }
    nstr[j] = -2;
    plen = j;
}

void solve()
{
    ans = -1;
    int id, mx = 0;
    for ( int i = 1; i < plen; i++ )
    {
        if(i < mx) 
            maxl[i] = Min(maxl[2*id-i], mx-i);
        else 
            maxl[i] = 1;
        
        while( nstr[i-maxl[i]]==nstr[i+maxl[i]] && nstr[i-maxl[i]]<=nstr[i-maxl[i]+2] )
            maxl[i]++;
        if(mx < i + maxl[i])
            { id = i; mx = i + maxl[i]; }

        ans = Max(ans, maxl[i]-1);
    }
}

int main()
{
    int all; cin >> all;
    while (all--)
    {
        cin >> N;
        for( int i = 0; i < N; i++ )
            scanf( "%d", &str[i] );
        init();
        solve();

        printf( "%d
", ans );
    }
    return 0;
}