「TJOI / HEOI2016」字符串

题目链接

问题分析

看见字符串,还涉及到子串和前缀的,当然就想到了SAM(其实我不会SA)。

首先我们把串反一下,就变成了求S[a..b]的所有子串与S[c..d]的最长后缀的长度。

我们把问题说成这样:求是S[a..b]子串的S[c..d]的最长后缀长度。于是我们就可以二分答案,然后判定是否在S[a..b]中出现就可以啦!

考虑二分答案后如何判定

假设我们现在判断Len是否可行。我们首先找到一个SAM上的状态,它的Right集合中含有d这个位置,并且Len介于这个状态的MinLen和MaxLen之间。那么只需要查询Right集合中有没有a+Len-1到b的元素即可。

具体细节是这样的:首先我们应该在建SAM的时候直接保存d最早出现的状态。这个状态就是Right集合中含有d,并且在Parent树上深度最深的节点。然后对于一个长度Len,我们在Parent树上倍增,找到[MinLen..MaxLen]包含Len的节点,判断的后缀就包含在这个状态里。同时,如果这个状态的Right集合中有[a+Len-1..b]的元素,那么说明这个串也存在于[a..b]中,否则就没有。

所以我们需要线段树合并求Right集合。然后就做完了!

时间复杂度\(O(n\log^2n)\)

参考程序

我不知道为什么写的这么翔

#include 
//#ifndef DEBUG
//  #define DEBUG
//#endif
using namespace std;

const int Maxn = 100010;
const int MaxLog = 20;
int n, m;
char Ch[ Maxn ];
int Index[ Maxn ];

namespace Seg { //线段树
    struct segmentTree {
        segmentTree *LeftChild, *RightChild;
        int Num;
        void Init() {
            Num = 0;
            LeftChild = RightChild = NULL;
        }
    };
    segmentTree *SegmentTree[ Maxn << 1 ];

    void Init() {
        for( int i = 0; i < Maxn << 1; ++i )
            SegmentTree[ i ] = NULL;
        return;
    }

    segmentTree *Insert( segmentTree *Index, int Left, int Right, int Key ) {
        if( Index == NULL ) {
            Index = new segmentTree;
            Index -> Init();
        }
        ++Index -> Num;
        if( Left == Right ) return Index;
        int Mid = ( Left + Right ) >> 1;
        if( Key <= Mid ) Index -> LeftChild = Insert( Index -> LeftChild, Left, Mid, Key );
        if( Key > Mid ) Index -> RightChild = Insert( Index -> RightChild, Mid + 1, Right, Key );
        return Index;
    }

    void Insert( int Pos, int Key ) {
#ifdef DEBUG
        printf( "Inform(Seg) : Add [ %d ] -> [ %d ]\n", Key, Pos );
#endif
        SegmentTree[ Pos ] = Insert( SegmentTree[ Pos ], 1, n, Key );
        return;
    }

    segmentTree *Merge( segmentTree *x, segmentTree *y, int Left, int Right ) {
        if( x == NULL && y == NULL ) return x;
        if( x != NULL && y == NULL ) return x;
        segmentTree *z = new segmentTree;//注意要新建节点
        if( x == NULL && y != NULL ) {
            *z = *y;
            return z;
        }
        *z = *x;
        z -> Num += y -> Num;
        if( Left == Right ) return z;
        int Mid = ( Left + Right ) >> 1;
        z -> LeftChild = Merge( z -> LeftChild, y -> LeftChild, Left, Mid );
        z -> RightChild = Merge( z -> RightChild, y -> RightChild, Mid + 1, Right );
        return z;
    }//线段树合并求Right集合

    void Play( segmentTree *Index, int Left, int Right ) {
        if( Index == NULL ) return;
        if( Index -> Num == 0 ) return;
        if( Left == Right ) {
            printf( "%d ", Left );
            return;
        }
        int Mid = ( Left + Right ) >> 1;
        Play( Index -> LeftChild, Left, Mid );
        Play( Index -> RightChild, Mid + 1, Right );
        return;
    }//Debug用

    void Merge( int x, int y ) {
#ifdef DEBUG
        printf( "Inform(Seg) : Merge %d <-- %d\n", x, y );
#endif
        SegmentTree[ x ] = Merge( SegmentTree[ x ], SegmentTree[ y ], 1, n );
#ifdef DEBUG
        printf( "  Right : " ); Play( SegmentTree[ x ], 1, n ); printf( "\n" );
#endif
        return;
    }//这是一个接口

    int Count( segmentTree *Index, int Left, int Right, int L, int R ) {
        if( Index == NULL ) return 0;
        if( L <= Left && Right <= R ) return Index -> Num;
        int Mid = ( Left + Right ) >> 1;
        int Ans = 0;
        if( L <= Mid ) Ans += Count( Index -> LeftChild, Left, Mid, L, R );
        if( R > Mid ) Ans += Count( Index -> RightChild, Mid + 1, Right, L, R );
        return Ans;
    }//求一段区间内的元素个数,用于判断区间内是否存在元素
} //Seg


namespace SAM {
    int Now, Used = 0, Last;
    struct suffixAutomaton {
        int Parent, Child[ 26 ], Len, MinLen;
    };
    suffixAutomaton SuffixAutomaton[ Maxn << 1 ];

    void Init() {
        Used = 0;
        memset( SuffixAutomaton, 0, sizeof( SuffixAutomaton ) );
        ++Used; Last = Used;
        return;
    }

    void Insert( int t ) {
        Now = ++Used;
        SuffixAutomaton[ Now ].Len = SuffixAutomaton[ Last ].Len + 1;
        int p = Last;
        for( ; p && !SuffixAutomaton[ p ].Child[ t ]; p = SuffixAutomaton[ p ].Parent )
            SuffixAutomaton[ p ].Child[ t ] = Now;
        Last = Now;
        if( !p ) {
            SuffixAutomaton[ Now ].Parent = 1;
            return;
        }
        int q = SuffixAutomaton[ p ].Child[ t ];
        if( SuffixAutomaton[ p ].Len + 1 == SuffixAutomaton[ q ].Len ) {
            SuffixAutomaton[ Now ].Parent = q;
            return;
        }
        int Clone = ++Used;
        SuffixAutomaton[ Clone ].Len = SuffixAutomaton[ p ].Len + 1;
        SuffixAutomaton[ Clone ].Parent = SuffixAutomaton[ q ].Parent;
        for( int i = 0; i < 26; ++i )
            SuffixAutomaton[ Clone ].Child[ i ] = SuffixAutomaton[ q ].Child[ i ];
        for( ; p && SuffixAutomaton[ p ].Child[ t ] == q; p = SuffixAutomaton[ p ].Parent )
            SuffixAutomaton[ p ].Child[ t ] = Clone;
        SuffixAutomaton[ Now ].Parent = SuffixAutomaton[ q ].Parent = Clone;
        return;
    }
} //SAM

namespace Tree {//维护Parent树
    struct edge {
        int To, Next;
    };
    edge Edge[ Maxn << 2 ];
    int Start[ Maxn << 1 ], Used = 0;
    int Deep[ Maxn << 1 ], D[ Maxn << 1 ][ MaxLog ];//倍增需要的数组

    inline void AddEdge( int x, int y ) {
#ifdef DEBUG
        printf( "Inform(Tree) : Link %d <---> %d\n", x, y );
#endif
        Edge[ ++Used ] = ( edge ) { y, Start[ x ] };
        Start[ x ] = Used;
        return;
    }

    void Build( int Index, int Father ) {//建树,顺便求出MinLen
        D[ Index ][ 0 ] = Father;
        Deep[ Index ] = Deep[ Father ] + 1;
        if( Index > 1 )
            SAM::SuffixAutomaton[ Index ].MinLen = SAM::SuffixAutomaton[ Father ].Len + 1;
        for( int i = 1; i < MaxLog; ++i ) 
            D[ Index ][ i ] = D[ D[ Index ][ i - 1 ] ][ i - 1 ];
        for( int t = Start[ Index ]; t; t = Edge[ t ].Next ) {
            int v = Edge[ t ].To;
            if( v == Father ) continue;
            Build( v, Index );
        }
        return;
    }

    void Init() {
        for( int i = 2; i <= SAM::Used; ++i ) {
            AddEdge( SAM::SuffixAutomaton[ i ].Parent, i );
            AddEdge( i, SAM::SuffixAutomaton[ i ].Parent );
        }
        Build( 1, 1 );
        return;
    }

    void Union( int Index, int Father ) {
        for( int t = Start[ Index ]; t; t = Edge[ t ].Next ) {
            int v = Edge[ t ].To;
            if( v == Father) continue;
            Union( v, Index );
            Seg::Merge( Index, v );
        }
        return;
    }//求Right集合

    void Union() {
        Union( 1, 1 );
        return;
    }//这是一个接口

    void Play( int Index, int Father ) {
        printf( "Index = %d, Father = %d\n", Index, Father );
        printf( "  Min = %d, Max = %d\n", SAM::SuffixAutomaton[ Index ].MinLen, SAM::SuffixAutomaton[ Index ].Len );
        printf( "  Right : " );
        Seg::Play( Seg::SegmentTree[ Index ], 1, n );
        printf( "\n" );
        for( int t = Start[ Index ]; t; t = Edge[ t ].Next ) {
            int v = Edge[ t ].To;
            if( v == Father ) continue;
            Play( v, Index );
        }
        return;
    }//Debug用
} //Tree

void Debug() {
    printf( "Total Used Point in SAM : %d\n", SAM::Used );
    Tree::Play( 1, 0 );
    printf( "Start : " );
    for( int i = 1; i <= n; ++i ) printf( "%d ", Index[ i ] );
    printf( "\n\n\n" );
    system( "pause" );
    return;
}

void Read() {
    scanf( "%d%d", &n, &m );
    scanf( "%s", Ch + 1 );
    for( int i = 1; i + i <= n; ++i ) 
        swap( Ch[ i ], Ch[ n - i + 1 ] );
#ifdef DEBUG
    printf( "%d %d\n", n, m );
    printf( "%s\n", Ch + 1 );
    system( "pause" );
#endif
    return;
}

void Build() {
    Seg::Init();
    SAM::Init();
    for( int i = 1; i <= n; ++i ) {
        SAM::Insert( Ch[ i ] - 'a' );
        Index[ i ] = SAM::Now;//存下Right集合包含位置i并且在Parent树中深度最深的点
        Seg::Insert( SAM::Now, i );//插入位置i
    }
    Tree::Init();
#ifdef DEBUG
    Debug();
#endif
    Tree::Union();
#ifdef DEBUG
    Debug();
#endif
    return;
}

bool Check( int Len, int Index, int Left, int Right ) {
#ifdef DEBUG
    printf( "Inform(Check) : Len = %d, Left = %d, Right = %d, Index = %d\n", Len, Left, Right, Index );
    printf( "Inform(Check) : Min = %d, Max = %d\n", SAM::SuffixAutomaton[ Index ].MinLen, SAM::SuffixAutomaton[ Index ].Len );
#endif
    if( !Len ) return true;
    if( Len > SAM::SuffixAutomaton[ Index ].Len ) return false;
    if( SAM::SuffixAutomaton[ Index ].MinLen <= Len && Len <= SAM::SuffixAutomaton[ Index ].Len ) {
#ifdef DEBUG
        printf( "Count = %d\n", Seg::Count( Seg::SegmentTree[ Index ], 1, n, Left, Right ) );
#endif
        return Seg::Count( Seg::SegmentTree[ Index ], 1, n, Left, Right );
    }
    for( int i = MaxLog - 1; i >= 0; --i ) //倍增
        if( SAM::SuffixAutomaton[ Tree::D[ Index ][ i ] ].MinLen > Len )
            Index = Tree::D[ Index ][ i ];
    Index = Tree::D[ Index ][ 0 ];
#ifdef DEBUG
    printf( "Inform(Check) : Min = %d, Max = %d\n", SAM::SuffixAutomaton[ Index ].MinLen, SAM::SuffixAutomaton[ Index ].Len );
    printf( "Count = %d\n", Seg::Count( Seg::SegmentTree[ Index ], 1, n, Left, Right ) );
#endif
    return Seg::Count( Seg::SegmentTree[ Index ], 1, n, Left, Right );
}

void Solve() {
    for( int i = 1; i <= m; ++i ) {
        int a, b, c, d;
        scanf( "%d%d%d%d", &a, &b, &c, &d );
        a = n - a + 1; b = n - b + 1; c = n - c + 1; d = n - d + 1;
        swap( a, b ); swap( c, d );
        int Left = 0, Right = d - c + 1;
        while( Left < Right ) {//二分答案
            int Mid = ( Left + Right + 1 ) >> 1;
            if( Check( Mid, Index[ d ], a + Mid - 1, b ) ) Left = Mid; else Right = Mid - 1;
        }
        printf( "%d\n", Left );
    }
    return;
}

int main() {
    Read();
    Build();
    Solve();
    return 0;
}

转载于:https://www.cnblogs.com/chy-2003/p/10512647.html

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