先按時(shí)間排序( 開始結(jié)束都可以 ) , 然后 dp( i ) = max( dp( i ) , dp( j ) + 1 ) ( j < i && 節(jié)日 j 結(jié)束時(shí)間在節(jié)日 i 開始時(shí)間之前 ) 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 參加節(jié)日慶祝

Time Limit:?5 Sec??Memory Limit:?64 MB
Submit:?262??Solved:?190
[Submit][Status][Discuss]

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.

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有N個(gè)節(jié)日每個(gè)節(jié)日有個(gè)開始時(shí)間,及持續(xù)時(shí)間. 牛想盡可能多的參加節(jié)日,問最多可以參加多少. 注意牛的轉(zhuǎn)移速度是極快的,不花時(shí)間.

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---------這個(gè)是時(shí)間軸.
--------------------
111111 2222223333344
55555555 777777 666

這個(gè)圖中1代表第一個(gè)節(jié)日從1開始,持續(xù)6個(gè)時(shí)間,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

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轉(zhuǎn)載于:https://www.cnblogs.com/JSZX11556/p/4558250.html

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