HDU 3577 Fast Arrangement ( 线段树 成段更新 区间最值 区间最大覆盖次数 )

线段树成段更新+区间最值。

注意某人的乘车区间是[a, b-1]，因为他在b站就下车了。

#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 = 1000100;

struct node
{
    int l, r;
};

node D[MAXN];
int limit, Q;
int maxi[ MAXN << 2 ];
int lazy[ MAXN << 2 ];
int N;

void build( int l, int r, int rt )
{
    maxi[rt] = lazy[rt] = 0;
    if ( l == r ) return;
    int m = ( l + r ) >> 1;
    build( lson );
    build( rson );
    return;
}

void PushUp( int rt )
{
    maxi[rt] = max( maxi[lc], maxi[rc] );
    return;
}

void PushDown( int rt )
{
    if ( lazy[rt] )
    {
        lazy[lc] += lazy[rt];
        lazy[rc] += lazy[rt];
        maxi[lc] += lazy[rt];
        maxi[rc] += lazy[rt];
        lazy[rt] = 0;
    }
    return;
}

void update( int L, int R, int l, int r, int rt )
{
    if ( L <= l && r <= R )
    {
        lazy[rt] += 1;
        maxi[rt] += 1;
        return;
    }
    PushDown( rt );
    int m = ( l + r ) >> 1;
    if ( L <= m ) update( L, R, lson );
    if ( R > m )  update( L, R, rson );
    PushUp( rt );
    return;
}

int query( int L, int R, int l, int r, int rt )
{
    if ( L <= l && r <= R )
    {
        return maxi[rt];
    }
    PushDown( rt );
    int m = ( l + r ) >> 1;

    int res = -10;
    if ( L <= m ) res = max( res, query( L, R, lson ) );
    if ( R > m )  res = max( res, query( L, R, rson ) );
    PushUp( rt );
    return res;
}

int main()
{
    int T, cas = 0;
    scanf( "%d", &T );
    while ( T-- )
    {
        scanf( "%d%d", &limit, &Q );
        N = 0;
        for ( int i = 0; i < Q; ++i )
        {
            int u, v;
            scanf( "%d%d", &u, &v );
            N = max( N, v );
            --v;
            D[i].l = u, D[i].r = v;
        }

        build( 1, N, 1 );
        printf( "Case %d:
", ++cas );
        for ( int i = 0; i < Q; ++i )
        {
            int ans = query( D[i].l, D[i].r, 1, N, 1 );
            if ( ans < limit )
            {
                printf( "%d ", i + 1 );
                update( D[i].l, D[i].r, 1, N, 1 );
            }
        }
        puts("
");
    }
    return 0;
}