HDU 4107 Gangster（线段树 特殊懒惰标记）

两种做法。

第一种：标记区间最大值和最小值，若区间最小值>=P，则本区间+2c，若区间最大值<P，则本区间+c。非常简单的区间更新。

最后发一点牢骚：最后query查一遍就行，我这个2B竟然写了个for循环每个点查了一遍orz……然后比赛的时候就一直TLE还查不出原因……感觉线段树对我就像个诅咒一样，每场必不出，不管是多么简单的线段树，都会错在千奇百怪的地方……说到底也不过是对线段树掌握的不扎实罢了，sigh……以后要多加练习！

#include <cstdio>

#include <cstring>

#include <cstdlib>



#define lson l, m, rt << 1

#define rson m + 1, r, rt << 1 | 1

#define lc rt << 1

#define rc rt << 1 | 1



using namespace std;



const int MAXN = 200100;



int N, M, P;

int sum[MAXN << 2];

int maxi[MAXN << 2];

int mini[MAXN << 2];

int lazy[MAXN << 2];



void build( int l, int r, int rt )

{

    sum[rt] = lazy[rt] = 0;

    maxi[rt] = 0;

    mini[rt] = 0;

    if ( l == r ) return;

    int m = ( l + r ) >> 1;

    build( lson );

    build( rson );

    return;

}



inline void PushDown( int rt, int m )

{

    if ( lazy[rt] )

    {

        lazy[lc] += lazy[rt];

        lazy[rc] += lazy[rt];

        sum[lc] += lazy[rt]*(m - (m >> 1) );

        sum[rc] += lazy[rt]*(m >> 1);

        maxi[lc] += lazy[rt], mini[lc] += lazy[rt];

        maxi[rc] += lazy[rt], mini[rc] += lazy[rt];

        lazy[rt] = 0;

    }

    return;

}



inline void PushUp( int rt )

{

    sum[rt] = sum[lc] + sum[rc];

    maxi[rt] = maxi[lc] > maxi[rc] ? maxi[lc] : maxi[rc];

    mini[rt] = mini[lc] < mini[rc] ? mini[lc] : mini[rc];

    return;

}



inline void update( int L, int R, int val, int l, int r, int rt )

{

    if ( L <= l && r <= R )

    {

        if ( maxi[rt] < P )

        {

            lazy[rt] += val;

            sum[rt] += val*(r - l + 1);

            maxi[rt] += val;

            mini[rt] += val;

            return;

        }

        else if ( mini[rt] >= P )

        {

            lazy[rt] += 2*val;

            sum[rt] += 2*val*(r - l + 1);

            maxi[rt] += 2*val;

            mini[rt] += 2*val;

            return;

        }

    }

    if ( l == r ) return;

    PushDown( rt, r - l + 1 );



    int m = ( l + r ) >> 1;

    if ( L <= m ) update( L, R, val, lson );

    if ( R > m )  update( L, R, val, rson );

    PushUp( rt );



    return;

}



bool first;



void query( int l, int r, int rt )

{

    if ( l == r )

    {

        if ( first ) putchar(' ');

        first = true;

        printf( "%d", sum[rt] );

        return;

    }

    PushDown( rt, r - l + 1 );

    int m = ( l + r ) >> 1;

    query( lson );

    query( rson );

    return;

}



int main()

{

    while ( scanf( "%d%d%d", &N, &M, &P ) == 3 )

    {

        build( 1, N, 1 );

        for ( int i = 0; i < M; ++i )

        {

            int a, b, c;

            scanf( "%d%d%d", &a, &b, &c );

            update( a, b, c, 1, N, 1 );

        }

        first = false;

        query( 1, N, 1 );

        puts("");

    }

    return 0;

}

第二种做法：线段树的特殊懒惰标记，方法跟 HDU 3954 一样。代码稍微改改就行。

#include <cstdio>

#include <cstring>

#include <cstdlib>

#include <algorithm>



#define lson l, m, rt << 1

#define rson m + 1, r, rt << 1 | 1

#define lc rt << 1

#define rc rt << 1 | 1



using namespace std;



const int MAXN = 200022;

const int INF = 1 << 30;



struct node

{

    int exp, level;

    int min_dis;

    int flag;

};



int N, M, P;

int K;

node Tr[ MAXN << 2 ];

int sum[20];



void build( int l, int r, int rt )

{

    Tr[rt].exp = Tr[rt].flag = 0;

    Tr[rt].level = 1;

    Tr[rt].min_dis = sum[1];

    if ( l == r ) return ;

    int m = ( l + r ) >> 1;

    build( lson );

    build( rson );

    return;

}



void PushDown( int rt )

{

    if ( Tr[rt].flag )

    {

        Tr[lc].exp += Tr[rt].flag * Tr[lc].level;

        Tr[lc].min_dis -= Tr[rt].flag;

        Tr[lc].flag += Tr[rt].flag;



        Tr[rc].exp += Tr[rt].flag * Tr[rc].level;

        Tr[rc].min_dis -= Tr[rt].flag;

        Tr[rc].flag += Tr[rt].flag;



        Tr[rt].flag = 0;

    }

    return;

}



void PushUp( int rt )

{

    Tr[rt].level = max( Tr[lc].level, Tr[rc].level );

    Tr[rt].exp = max( Tr[lc].exp, Tr[rc].exp );

    Tr[rt].min_dis = min( Tr[lc].min_dis, Tr[rc].min_dis );

    return;

}



void Update( int L, int R, int v, int l, int r, int rt )

{

    if ( l == r )

    {

        Tr[rt].exp += Tr[rt].level * v;

        while ( Tr[rt].exp >= sum[ Tr[rt].level ] )

            ++Tr[rt].level;

        Tr[rt].min_dis = ( sum[ Tr[rt].level ] - Tr[rt].exp ) / Tr[rt].level;

        if( ( sum[ Tr[rt].level ] - Tr[rt].exp ) % Tr[rt].level ) ++Tr[rt].min_dis;

        return;

    }

    int m = ( l + r ) >> 1;



    if ( L == l && r == R )

    {

        if ( v >= Tr[rt].min_dis )

        {

            PushDown(rt);

            if ( R <= m ) Update( L, R, v, lson );

            else if ( L > m ) Update( L, R, v, rson );

            else

            {

                Update( L, m, v, lson );

                Update( m + 1, R, v, rson );

            }

            PushUp(rt);

        }

        else

        {

            Tr[rt].exp += Tr[rt].level * v;

            Tr[rt].min_dis -= v;

            Tr[rt].flag += v;

        }

        return;

    }



    PushDown(rt);



    if ( R <= m ) Update( L, R, v, lson );

    else if ( L > m ) Update( L, R, v, rson );

    else

    {

        Update( L, m, v, lson );

        Update( m + 1, R, v, rson );

    }



    PushUp(rt);



    return;

}



bool first;



void Query( int l, int r, int rt )

{

    if ( l == r )

    {

        if ( first ) putchar(' ');

        first = true;

        printf( "%d", Tr[rt].exp );

        return;

    }

    PushDown(rt);

    int m = ( l + r ) >> 1;

    Query( lson );

    Query( rson );

    return;

}



int main()

{

    K = 2;

    while ( scanf( "%d%d%d", &N, &M, &P ) == 3 )

    {

        sum[1] = P;

        sum[2] = INF;



        build( 1, N, 1 );

        while ( M-- )

        {

            int a, b, c;

            scanf( "%d%d%d", &a, &b, &c );

            Update( a, b, c, 1, N, 1 );

        }

        first = false;

        Query( 1, N, 1 );

        puts("");

    }

    return 0;

}