[USACO12DEC]First! G

luogu3065

考虑每一个字符串成为答案的可能

这意味着从字典树根到字符串最后一位就恰好对应重新定义的字典序

在第$i$层的时候，想要走特定点，意味着这个点的字典序必须大于该层隶属于同一个父亲的所有点(兄弟)

建立$u\to v$的有向边，表示$u$的字典序$<v$

最后肯定字典序不能出现环

拓扑判环

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 300005
queue < int > q;
int n, cnt, tot;
string s[maxn];
bool tag[maxn], vis[maxn], in[26];
int d[26];
bool E[26][26];
int trie[maxn][26];void insert( string s ) {int now = 0;int len = s.length();for( int i = 0;i < len;i ++ ) {int son = s[i] - 'a';if( ! trie[now][son] ) trie[now][son] = ++ cnt;now = trie[now][son];}tag[now] = 1;
}bool check( string s ) {memset( d, 0, sizeof( d ) );memset( E, 0, sizeof( E ) );memset( in, 0, sizeof( in ) );int len = s.length();int now = 0;for( int i = 0;i < len;i ++ ) {if( tag[now] ) return 0;int son = s[i] - 'a';for( int j = 0;j < 26;j ++ )if( son ^ j and ! E[son][j] and trie[now][j] ) {E[son][j] = 1;d[j] ++;}now = trie[now][son];}for( int i = 0;i < 26;i ++ )if( ! d[i] ) q.push( i );while( ! q.empty() ) {int now = q.front(); q.pop();in[now] = 1;for( int i = 0;i < 26;i ++ )if( E[now][i] ) {d[i] --;if( ! d[i] ) q.push( i );}}for( int i = 0;i < 26;i ++ )if( ! in[i] ) return 0;return 1;
}int main() {scanf( "%d", &n );for( int i = 1;i <= n;i ++ ) {cin >> s[i];insert( s[i] );}int ans = 0;for( int i = 1;i <= n;i ++ )if( check( s[i] ) ) vis[i] = 1, ans ++;printf( "%d\n", ans );for( int i = 1;i <= n;i ++ )if( vis[i] ) cout << s[i] << endl;return 0;
}

[JSOI2009]电子字典

luogu4407

由于一个字符串长度不超过$20$，且要求相异必须为$1$

其实状态数是可以直接爆搜的

#include <map>
#include <cstdio>
#include <cstring>
using namespace std;
#define maxs 25
#define maxn 200005
map < int, int > mp;
int n, m, ans, cnt, len;
char s[maxn];
bool tag[maxn];
int trie[maxn][26];void insert() {int now = 0;len = strlen( s + 1 );for( int i = 1;i <= len;i ++ ) {int c = s[i] - 'a';if( ! trie[now][c] ) trie[now][c] = ++ cnt;now = trie[now][c];}tag[now] = 1;
}bool find() {int now = 0;len = strlen( s + 1 );for( int i = 1;i <= len;i ++ ) {int c = s[i] - 'a';if( ! trie[now][c] ) return 0;now = trie[now][c];}return tag[now];
}void dfs( int ip, int now, bool k ) {if( ip > len ) {if( k ) {ans += ( tag[now] and ! mp[now] );mp[now] = 1;}else for( int i = 0;i < 26;i ++ )if( ! mp[trie[now][i]] and tag[trie[now][i]] )ans ++, mp[trie[now][i]] = 1;return;}int c = s[ip] - 'a';if( k ) {if( ! trie[now][c] ) return;else dfs( ip + 1, trie[now][c], 1 );}else {if( trie[now][c] ) dfs( ip + 1, trie[now][c], 0 );dfs( ip + 1, now, 1 );for( int i = 0;i < 26;i ++ )if( trie[now][i] ) {dfs( ip, trie[now][i], 1 );dfs( ip + 1, trie[now][i], 1 );}}
}int main() {scanf( "%d %d", &n, &m );for( int i = 1;i <= n;i ++ ) {scanf( "%s", s + 1 );insert();}for( int i = 1;i <= m;i ++ ) {scanf( "%s", s + 1 );if( find() ) printf( "-1\n" );else {dfs( 1, 0, 0 );printf( "%d\n", ans );ans = 0;mp.clear();}}return 0;
}

[IOI2008] Type Printer

luogu4683

建立字典树

贪心的最后剩的字符越多越好

如果这个串不是最后一个，那么这个串肯定会原路返回直到遇到分叉点，走兄弟点

所以求出每个串与最近分叉点的距离，选最大值

然后把这个串一路上的点打上最后访问标记

接着开始遍历，先访问未标记点，再访问标记点，都清空打印机

最后把直到最后一个打印P为止的所有-操作都扔出操作顺序集合

#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
using namespace std;
#define maxn 500005
vector < char > ans;
vector < pair < int, int > > G;
int n, cnt;
char s[25];
bool tag[maxn], vis[maxn];
int dep[maxn], son[maxn];
int trie[maxn][26];void insert() {int len = strlen( s + 1 ), now = 0;for( int i = 1;i <= len;i ++ ) {int c = s[i] - 'a';if( ! trie[now][c] ) trie[now][c] = ++ cnt;dep[trie[now][c]] = dep[now] + 1;now = trie[now][c];}tag[now] = 1;
}void dfs1( int now, int lst ) {if( tag[now] ) G.push_back( make_pair( now, max( lst, 0 ) ) );int tot = 0;for( int i = 0;i < 26;i ++ )if( trie[now][i] ) tot ++;for( int i = 0;i < 26;i ++ )if( trie[now][i] )dfs1( trie[now][i], lst == -1 ? ( tot > 1 ? now : lst ) : lst );
}void dfs2( int now, int End, bool &flag ) {if( now == End ) { flag = 1; return; }for( int i = 0;i < 26;i ++ )if( trie[now][i] ) {vis[trie[now][i]] = 1;dfs2( trie[now][i], End, flag );if( flag ) return;vis[trie[now][i]] = 0;}
}void dfs3( int now ) {if( tag[now] ) ans.push_back( 'P' );for( int i = 0;i < 26;i ++ ) {if( trie[now][i] and ! vis[trie[now][i]] ) {ans.push_back( i + 'a' );dfs3( trie[now][i] );ans.push_back( '-' );}}for( int i = 0;i < 26;i ++ )if( vis[trie[now][i]] ) {ans.push_back( i + 'a' );dfs3( trie[now][i] );ans.push_back( '-' );}
}int main() {memset( son, -1, sizeof( son ) );scanf( "%d", &n );for( int i = 1;i <= n;i ++ ) {scanf( "%s", s + 1 );insert();}dfs1( 0, -1 );int Dep = 0, pos;for( int i = 0;i < G.size();i ++ ) {if( dep[G[i].first] - dep[G[i].second] > Dep ) {Dep = dep[G[i].first] - dep[G[i].second];pos = G[i].first;}}bool flag = 0;dfs2( 0, pos, flag );dfs3( 0 );for( int i = ans.size() - 1;~ i;i -- )if( ans[i] != 'P' ) continue;else { cnt = i + 1; break; }printf( "%d\n", cnt );for( int i = 0;i < cnt;i ++ )printf( "%c\n", ans[i] );return 0;
}

Nikitosh and xor

CodeChef

${⨁}_{i=l}^r{a}_i\to b_r⨁b_{l-1}$

枚举$i$，利用字典树，边插入边询问，将$b_i$的二进制位从高到低放入字典树

求出$b_i$与前面某个$b_j$的异或最大值

同理，从后往前操作一样

${⨁}_{i=l}^r{a}_i\to c_l⨁c_{r+1}$

枚举$i$，利用字典树，边插入边询问，将$b_i$的二进制位从高到低放入字典树

求出$b_i$与后面某个$b_j$的异或最大值

然后前缀/后缀max递推

枚举断点$i$，直接利用$pr{e}_i+su{f}_{i+1}$更新答案

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define int long long
#define maxn 400005
int trie[maxn * 30][2];
int a[maxn], b[maxn], c[maxn], pre[maxn], suf[maxn];
int n, cnt;void insert( int x ) {int now = 0;for( int i = 29;~ i;i -- ) {int k = x >> i & 1;if( ! trie[now][k] ) trie[now][k] = ++ cnt;now = trie[now][k];}
}int query( int x ) {int now = 0, ans = 0;for( int i = 29;~ i;i -- ) {int k = x >> i & 1;if( trie[now][k ^ 1] )ans += 1 << i, now = trie[now][k ^ 1];elsenow = trie[now][k];}return ans;
}signed main() {scanf( "%lld", &n );for( int i = 1;i <= n;i ++ )scanf( "%lld", &a[i] );for( int i = 1;i <= n;i ++ )b[i] = a[i] ^ b[i - 1];for( int i = n;i;i -- )c[i] = c[i + 1] ^ a[i];insert( 0 );for( int i = 1;i <= n;i ++ ) {pre[i] = max( pre[i - 1], query( b[i] ) );insert( b[i] );}memset( trie, 0, sizeof( trie ) );cnt = 0;insert( 0 );for( int i = n;i;i -- ) {suf[i] = max( suf[i + 1], query( c[i] ) );insert( c[i] );}int ans = 0;for( int i = 1;i < n;i ++ )ans = max( ans, pre[i] + suf[i + 1] );printf( "%lld\n", ans );return 0;
}