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 h…

将区间按左端点排序. f(i)=max{f(j)+1}(p[j].x+p[j].y<=p[i].x && j<i) #include<cstdio> #include<algorithm> using namespace std; int n,f[10001]; struct Point{int x,y;}p[10001]; bool operator < (const Point &a,const Point &b){return…

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 ) -------------------------------------------------------------------------------- #include<cstdio&g…

1664: [Usaco2006 Open]County Fair Events 参加节日庆祝 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 255  Solved: 185[Submit][Status][Discuss] Description Farmer John has returned to the County Fair so he can attend the special events (concerts, rodeos, coo…

http://www.lydsy.com/JudgeOnline/problem.php?id=1664 和之前的那题一样啊.. 只不过权值变为了1.. 同样用线段树维护区间,然后在区间范围内dp. upd:(其实权值为1的可以直接贪心....右端点来就行了... #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream>…

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 h…

把长度转成右端点,按右端点排升序,f[i]=max(f[j]&&r[j]<l[i]),因为r是有序的,所以可以直接二分出能转移的区间(1,w),然后用树状数组维护区间f的max,每次转移的时候直接从树状数组上查询前缀max即可,然后把更新出来的f[i]update进树状数组 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N…

先按照结束时间进行排序,取第一个节日的结束时间作为当前时间,然后从第二个节日开始搜索,如果下一个节日的开始时间大于当前的时间,那么就参加这个节日,并更新当前时间 #include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 1e4+5; struct node { int be, ed; // 开始时间和结束时间 }nds[maxn]; bool cmp(node x, node y)…

Search GO 说明:输入题号直接进入相应题目,如需搜索含数字的题目,请在关键词前加单引号 Problem ID Title Source AC Submit Y 1000 A+B Problem 10983 18765 Y 1036 [ZJOI2008]树的统计Count 5293 13132 Y 1588 [HNOI2002]营业额统计 5056 13607 1001 [BeiJing2006]狼抓兔子 4526 18386 Y 2002 [Hnoi2010]Bounce 弹飞绵羊 43…

bzoj上的usaco题目还是很好的(我被虐的很惨. 有必要总结整理一下. 1592: [Usaco2008 Feb]Making the Grade 路面修整 一开始没有想到离散化.然后离散化之后就很好做了.F[I,j]表示第i个点,高度>=j或<=j,f[I,j]=min(f[i-1,j]+abs(b[j]-a[i]),f[I,j-1]) 1593: [Usaco2008 Feb]Hotel 旅馆 线段树 ★1594: [Usaco2008 Jan]猜数游戏 二分答案然后写线段树维护 15…