bzoj3172: [Tjoi2013]单词

Description
某人读论文，一篇论文是由许多单词组成。但他发现一个单词会在论文中出现很多次，现在想知道每个单词分别在论文中出现多少次。

Input
第一个一个整数N,表示有多少个单词，接下来N行每行一个单词。每个单词由小写字母组成，N<=200,单词长度不超过10^6

Output
输出N个整数，第i行的数字表示第i个单词在文章中出现了多少次。

Sample Input
3
a
aa
aaa
Sample Output
6
3
1

题意差评，没写到底是每个字符串长10^6还是总长10^6..

这道题有很多做法：

1）用AC自动机嘛.. 就是统计一个子树和就好了（口胡）
2）用SAM做嘛.. 只要统计一个Right数组就可以了啊..
3）SAM建后缀树，这个请见blog：http://blog.csdn.net/werkeytom_ftd/article/details/51154465
这个不是我的，但我很快会去学..

然后这道题对多串有两种处理方法：（1）中间加一个字符 （2）广义SAM

插字符：

#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; const int Maxn = 2500010; int F[Maxn], d[Maxn], ch[Maxn][27], tot, now; char s[Maxn]; int len, n; int Rsort[Maxn], rk[Maxn], G[Maxn]; int copy ( int p, int c ){     int x = ++tot, y = ch[p][c];     d[x] = d[p]+1;     for ( int i = 0; i < 27; i ++ ) ch[x][i] = ch[y][i];     F[x] = F[y]; F[y] = x;     while ( ~p && ch[p][c] == y ){ ch[p][c] = x; p = F[p]; }     return x; } void add ( int c ){     int p, o;     if ( p = ch[now][c] ){         if ( d[p] != d[now]+1 ) copy ( now, c );         now = ch[now][c];     }     else {         d[o=++tot] = d[now]+1; p = now; now = o; G[o] = 1;         while ( ~p && !ch[p][c] ){ ch[p][c] = o; p = F[p]; }         F[o] = ~p ? ( d[p]+1 == d[ch[p][c]] ? ch[p][c] : copy ( p, c ) ) : 0;     } } int main (){     int i, j, k;     scanf ( "%d", &n );     len = 0;     for ( i = 1; i <= n; i ++ ){         scanf ( "%s", s+len );         len = strlen (s);         s[len++] = 'a'+26;     }     F[0] = -1;     for ( i = 0; i < len; i ++ ){         add (s[i]-'a');     }     for ( i = 1; i <= tot; i ++ ) Rsort[d[i]] ++;     for ( i = 1; i <= len; i ++ ) Rsort[i] += Rsort[i-1];     for ( i = tot; i >= 1; i -- ) rk[Rsort[d[i]]--] = i;     for ( i = tot; i >= 1; i -- ){         G[F[rk[i]]] += G[rk[i]];     }     j = 0;     for ( i = 1; i <= n; i ++ ){         if ( s[j] == 'a'+26 ) j ++;         now = 0;         while ( s[j] != 'a'+26 ){             int c = s[j]-'a';             now = ch[now][c];             j ++;         }         printf ( "%d\n", G[now] );     }     return 0; }

广义SAM：

#include <cstdio> #include <cstring> #include <cstdlib> #include <algorithm> using namespace std; const int Maxn = 2500010; int F[Maxn], d[Maxn], ch[Maxn][26], tot, now; char s[Maxn]; int len, n; int Rsort[Maxn], rk[Maxn], G[Maxn]; int copy ( int p, int c ){     int x = ++tot, y = ch[p][c];     d[x] = d[p]+1;     for ( int i = 0; i < 26; i ++ ) ch[x][i] = ch[y][i];     F[x] = F[y]; F[y] = x;     while ( ~p && ch[p][c] == y ){ ch[p][c] = x; p = F[p]; }     return x; } void add ( int c ){     int p, o;     if ( p = ch[now][c] ){         if ( d[p] != d[now]+1 ) copy ( now, c );         now = ch[now][c]; G[now] ++;     }     else {         d[o=++tot] = d[now]+1; p = now; now = o; G[o] = 1;         while ( ~p && !ch[p][c] ){ ch[p][c] = o; p = F[p]; }         F[o] = ~p ? ( d[p]+1 == d[ch[p][c]] ? ch[p][c] : copy ( p, c ) ) : 0;     } } int main (){     int i, j, k;     scanf ( "%d", &n );     len = 0;     for ( i = 1; i <= n; i ++ ){         scanf ( "%s", s+len );         len = strlen (s);         s[len++] = '$';     }     F[0] = -1;     for ( i = 0; i < len; i ++ ){         if ( s[i] != '$' ) add (s[i]-'a');         else now = 0;     }     for ( i = 1; i <= tot; i ++ ) Rsort[d[i]] ++;     for ( i = 1; i <= len; i ++ ) Rsort[i] += Rsort[i-1];     for ( i = tot; i >= 1; i -- ) rk[Rsort[d[i]]--] = i;     for ( i = tot; i >= 1; i -- ){         G[F[rk[i]]] += G[rk[i]];     }     j = 0;     for ( i = 1; i <= n; i ++ ){         if ( s[j] == '$' ) j ++;         now = 0;         while ( s[j] != '$' ){             int c = s[j]-'a';             now = ch[now][c];             j ++;         }         printf ( "%d\n", G[now] );     }     return 0; }

广义SAM：652ms..
插字符做法：1020ms..