BZOJ 1664: [Usaco2006 Open]County Fair Events 参加节日庆祝( dp )

先按时间排序( 开始结束都可以 ) , 然后 dp( i ) = max( dp( i ) , dp( j ) + 1 ) ( j < i && 节日 j 结束时间在节日 i 开始时间之前 ) answer = max( dp( i ) ) ( 1 <= i <= n )

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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 ) )

using namespace std;

const int maxn = 10000 + 5;

struct data {

int l , r;

void Read() {

scanf( "%d%d" , &l , &r );

r += l - 1;

}

bool operator < ( const data &rhs ) const {

return r < rhs.r;

}

};

data A[ maxn ];

int dp[ maxn ];

int main() {

// freopen( "test.in" , "r" , stdin );

int n;

cin >> n;

rep( i , n )

   A[ i ].Read();

sort( A , A + n );

rep( i , n ) {

dp[ i ] =1;

rep( j , i ) if( A[ j ].r < A[ i ].l )

   dp[ i ] = max( dp[ i ] , dp[ j ] + 1 );

}

int ans = 0;

rep( i , n ) 

   ans = max( ans , dp[ i ] );

cout << ans << "\n";

return 0;

}

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1664: [Usaco2006 Open]County Fair Events 参加节日庆祝

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 262  Solved: 190
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Description

Farmer John has returned to the County Fair so he can attend the special events (concerts, rodeos, cooking shows, etc.). He wants to attend as many of the N (1 <= N <= 10,000) special events as he possibly can. He's rented a bicycle so he can speed from one event to the next in absolutely no time at all (0 time units to go from one event to the next!). Given a list of the events that FJ might wish to attend, with their start times (1 <= T <= 100,000) and their durations (1 <= L <= 100,000), determine the maximum number of events that FJ can attend. FJ never leaves an event early.

有N个节日每个节日有个开始时间,及持续时间. 牛想尽可能多的参加节日,问最多可以参加多少. 注意牛的转移速度是极快的,不花时间.

Input

* Line 1: A single integer, N.

* Lines 2..N+1: Each line contains two space-separated integers, T and L, that describe an event that FJ might attend.

Output

* Line 1: A single integer that is the maximum number of events FJ can attend.

Sample Input

7
1 6
8 6
14 5
19 2
1 8
18 3
10 6

INPUT DETAILS:

Graphic picture of the schedule:
11111111112
12345678901234567890---------这个是时间轴.
--------------------
111111 2222223333344
55555555 777777 666

这个图中1代表第一个节日从1开始,持续6个时间,直到6.

Sample Output

4

OUTPUT DETAILS:

FJ can do no better than to attend events 1, 2, 3, and 4.

HINT

Source

Silver