文章目錄

• [USACO12DEC]First! G
• [JSOI2009]電子字典
• [IOI2008] Type Printer
• Nikitosh and xor

[USACO12DEC]First! G

luogu3065

考慮每一個(gè)字符串成為答案的可能

這意味著從字典樹(shù)根到字符串最后一位就恰好對(duì)應(yīng)重新定義的字典序

在第$i$層的時(shí)候，想要走特定點(diǎn)，意味著這個(gè)點(diǎn)的字典序必須大于該層隸屬于同一個(gè)父親的所有點(diǎn)(兄弟)

建立$u\to v$的有向邊，表示$u$的字典序$<v$

最后肯定字典序不能出現(xiàn)環(huán)

拓?fù)渑协h(huán)

#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]電子字典

luogu4407

由于一個(gè)字符串長(zhǎng)度不超過(guò)$20$，且要求相異必須為$1$

其實(shí)狀態(tài)數(shù)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

建立字典樹(shù)

貪心的最后剩的字符越多越好

如果這個(gè)串不是最后一個(gè)，那么這個(gè)串肯定會(huì)原路返回直到遇到分叉點(diǎn)，走兄弟點(diǎn)

所以求出每個(gè)串與最近分叉點(diǎn)的距離，選最大值

然后把這個(gè)串一路上的點(diǎn)打上最后訪問(wèn)標(biāo)記

接著開(kāi)始遍歷，先訪問(wèn)未標(biāo)記點(diǎn)，再訪問(wèn)標(biāo)記點(diǎn)，都清空打印機(jī)

最后把直到最后一個(gè)打印P為止的所有-操作都扔出操作順序集合

#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

${?}_{i=l}^r{a}_i\to b_r?b_{l?1}$

枚舉$i$，利用字典樹(shù)，邊插入邊詢(xún)問(wèn)，將$b_i$的二進(jìn)制位從高到低放入字典樹(shù)

求出$b_i$與前面某個(gè)$b_j$的異或最大值

同理，從后往前操作一樣

${?}_{i=l}^r{a}_i\to c_l?c_{r+1}$

枚舉$i$，利用字典樹(shù)，邊插入邊詢(xún)問(wèn)，將$b_i$的二進(jìn)制位從高到低放入字典樹(shù)

求出$b_i$與后面某個(gè)$b_j$的異或最大值

然后前綴/后綴max遞推

枚舉斷點(diǎn)$i$，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; }

總結(jié)

以上是生活随笔為你收集整理的数据结构之trie树——First! G，电子字典，Type Printer，Nikitosh and xor的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問(wèn)題。

如果覺(jué)得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。