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+;

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

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

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

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

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

         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]-);
     }
 }

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

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+;

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

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

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

 void solve()
 {
     ans = -;
     int id, mx = ;
     for ( int i = ; i < plen; i++ )
     {
         if(i < mx)
             maxl[i] = Min(maxl[*id-i], mx-i);
         else
             maxl[i] = ;

         while( nstr[i-maxl[i]]==nstr[i+maxl[i]] && nstr[i-maxl[i]]<=nstr[i-maxl[i]+] )
             maxl[i]++;
         if(mx < i + maxl[i])
             { id = i; mx = i + maxl[i]; }

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

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

         printf( "%d\n", ans );
     }
     return ;
 }