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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;
int Max , add;
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 );
o.Max += o.add;
}
int L , R;
void update( int x ) {
Node &o = tree[ x ];
if( L <= o.l && o.r <= R )
o.add += 1;
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
[Submit][Status][Discuss]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
有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求
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.
Sample Input
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.
HINT
不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3
Source
BZOJ 1651: [Usaco2006 Feb]Stall Reservations 专用牛棚( 线段树 )
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原文地址：http://www.cnblogs.com/JSZX11556/p/4557742.html
