# BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚( 线段树 )

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

#define rep( i , n ) for( int i = 0 ; i < n ; i++ )
#define clr( x , c ) memset( x , c , sizeof( x ) )
#define L( x ) ( x << 1 )
#define R( x ) ( L( x ) ^ 1 )
#define LC( x ) tree[ L( x ) ]
#define RC( x ) tree[ R( x ) ]
#define mid( l , r ) ( ( l + r ) >> 1 )

using namespace std;

const int n = 1000000;

struct Node {
int l , r;
Node() : Max( 0 ) , add( 0 ) { }
};

Node tree[ n << 2 ];

void maintain( int x ) {
Node &o = tree[ x ];
o.Max = 0;
if( o.r > o.l )
o.Max = max( LC( x ).Max , RC( x ).Max );

}

int L , R;

void update( int x ) {
Node &o = tree[ x ];
if( L <= o.l && o.r <= R )

else {
int m = mid( o.l , o.r );
if( L <= m ) update( L( x ) );
if( m < R ) update( R( x ) );
}
maintain( x );
}

void build( int x , int l , int r ) {
Node &o = tree[ x ];
o.l = l , o.r = r;
if( l == r )
return ;

int m = mid( l , r );
build( L( x ) , l , m );
build( R( x ) , m + 1 , r );
int main() {
// freopen( "test.in" , "r" , stdin );
int m;
cin >> m;
build( 1 , 1 , n );
while( m-- ) {
scanf( "%d%d" , &L , &R );
update( 1 );
}
printf( "%d\n" , tree[ 1 ].Max );
return 0;
}

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## 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec  Memory Limit: 64 MB
Submit: 587  Solved: 327
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## Description

Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows. Help FJ by determining: * The minimum number of stalls required in the barn so that each cow can have her private milking period * An assignment of cows to these stalls over time

## Input

* Line 1: A single integer, N

* Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.

## Output

* Line 1: The minimum number of stalls the barn must have.

* Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.

5
1 10
2 4
3 6
5 8
4 7

## Sample Output

4

OUTPUT DETAILS:

Here's a graphical schedule for this output:

Time 1 2 3 4 5 6 7 8 9 10
Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..
Stall 3 .. .. c3>>>>>>>>> .. .. .. ..
Stall 4 .. .. .. c5>>>>>>>>> .. .. ..

Other outputs using the same number of stalls are possible.

## Source

posted @ 2015-06-07 09:54  JSZX11556  阅读(136)  评论(0编辑  收藏  举报