模板，但是对这个算法还是不太清楚,真实不明觉厉....
1 #include <iostream> 2 #include <cstdio> 3 #include <string.h> 4 #pragma warning ( disable : 4996 ) 5 using namespace std; 6 7 inline int Max(int a,int b) { return a>b?a:b; } 8 inline int Min(int a,int b) { return a>b?b:a; } 9 const int inf = 0x3f3f3f3f; 10 const int maxn = 1e5+1e4+5; 11 12 char str[maxn]; 13 char nstr[maxn<<1]; 14 int maxlen[maxn<<1]; 15 int len, plen, ans; 16 17 void init() 18 { 19 memset( nstr, 0, sizeof(nstr) ); 20 memset( maxlen, 0, sizeof(maxlen) ); 21 len = strlen(str); 22 nstr[0] = '$'; nstr[1] = '#'; 23 24 int j = 2; 25 for ( int i = 0; i < len; i++ ) 26 { 27 nstr[j++] = str[i]; 28 nstr[j++] = '#'; 29 } 30 nstr[j] = ' '; 31 plen = j; 32 } 33 34 void solve() 35 { 36 ans = -1; 37 int id, mx = 0; 38 39 for ( int i = 1; i < plen; i++ ) 40 { 41 if( i < mx ) 42 maxlen[i] = Min( maxlen[2*id-i], mx-i ); 43 else 44 maxlen[i] = 1; 45 46 while( nstr[i-maxlen[i]] == nstr[i+maxlen[i]] ) 47 maxlen[i]++; 48 49 if ( mx < i + maxlen[i] ) 50 { 51 id = i; 52 mx = i + maxlen[i]; 53 } 54 ans = Max(ans, maxlen[i]-1); 55 } 56 } 57 58 int main() 59 { 60 while ( ~scanf("%s", str) ) 61 { 62 init(); 63 solve(); 64 printf( "%d ", ans ); 65 } 66 return 0; 67 }
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
1 #include <iostream> 2 #include <cstdio> 3 #include <string.h> 4 #pragma warning ( disable : 4996 ) 5 using namespace std; 6 7 inline int Max(int a,int b) { return a>b?a:b; } 8 inline int Min(int a,int b) { return a>b?b:a; } 9 const int inf = 0x3f3f3f3f; 10 const int maxn = 1e5+5; 11 12 int maxl[maxn<<1]; 13 int str[maxn], nstr[maxn<<1]; 14 int N, ans, plen; 15 16 void init() 17 { 18 memset( maxl, 0, sizeof(maxl) ); 19 memset( nstr, 0, sizeof(nstr) ); 20 nstr[0] = 0; nstr[1] = -1; 21 22 int j = 2; 23 for( int i = 0; i < N; i++ ) 24 { nstr[j++] = str[i]; nstr[j++] = -1; } 25 nstr[j] = -2; 26 plen = j; 27 } 28 29 void solve() 30 { 31 ans = -1; 32 int id, mx = 0; 33 for ( int i = 1; i < plen; i++ ) 34 { 35 if(i < mx) 36 maxl[i] = Min(maxl[2*id-i], mx-i); 37 else 38 maxl[i] = 1; 39 40 while( nstr[i-maxl[i]]==nstr[i+maxl[i]] && nstr[i-maxl[i]]<=nstr[i-maxl[i]+2] ) 41 maxl[i]++; 42 if(mx < i + maxl[i]) 43 { id = i; mx = i + maxl[i]; } 44 45 ans = Max(ans, maxl[i]-1); 46 } 47 } 48 49 int main() 50 { 51 int all; cin >> all; 52 while (all--) 53 { 54 cin >> N; 55 for( int i = 0; i < N; i++ ) 56 scanf( "%d", &str[i] ); 57 init(); 58 solve(); 59 60 printf( "%d ", ans ); 61 } 62 return 0; 63 }