题目 3389: [Usaco2004 Dec]Cleaning Shifts安排值班 Time Limit: 1 Sec  Memory Limit: 128 MB Description     一天有T(1≤T≤10^6)个时段．约翰正打算安排他的N(1≤N≤25000)只奶牛来值班,打扫 打扫牛棚卫生．每只奶牛都有自己的空闲时间段[Si,Ei](1≤Si≤Ei≤T),只能把空闲的奶牛安排出来值班．而且,每个时间段必需有奶牛在值班．  那么,最少需要动用多少奶牛参与值班呢?如果没有办法安排…

3389: [Usaco2004 Dec]Cleaning Shifts安排值班 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 218  Solved: 86[Submit][Status][Discuss] Description     一天有T(1≤T≤10^6)个时段．约翰正打算安排他的N(1≤N≤25000)只奶牛来值班,打扫 打扫牛棚卫生．每只奶牛都有自己的空闲时间段[Si,Ei](1≤Si≤Ei≤T),只能把空闲的奶牛安排出来值班．而…

3389: [Usaco2004 Dec]Cleaning Shifts安排值班 Time Limit: 1 Sec  Memory Limit: 128 MB Description     一天有T(1≤T≤10^6)个时段．约翰正打算安排他的N(1≤N≤25000)只奶牛来值班,打扫 打扫牛棚卫生．每只奶牛都有自己的空闲时间段[Si,Ei](1≤Si≤Ei≤T),只能把空闲的奶牛安排出来值班．而且,每个时间段必需有奶牛在值班．  那么,最少需要动用多少奶牛参与值班呢?如果没有办法安排出合理…

3389: [Usaco2004 Dec]Cleaning Shifts安排值班 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 102  Solved: 46[Submit][Status][Discuss] Description     一天有T(1≤T≤10^6)个时段．约翰正打算安排他的N(1≤N≤25000)只奶牛来值班,打扫 打扫牛棚卫生．每只奶牛都有自己的空闲时间段[Si,Ei](1≤Si≤Ei≤T),只能把空闲的奶牛安排出来值班．而…

http://www.lydsy.com/JudgeOnline/problem.php?id=3389 显然左端点排序后,依次取. 要考虑下一次取的方案: 待选点为a[j].x<=a[now].y+1的所有点j,其中now是当前所选 那么我们要在这些点内做决策 贪心就是取y最大的待选点,即 max(a[j].y)的j 实现中细节太多QAQ,写了挺久的.. #include <cstdio> #include <cstring> #include <cmath>…

3389: [Usaco2004 Dec]Cleaning Shifts安排值班 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 45  Solved: 26[Submit][Status] Description     一天有T(1≤T≤10^6)个时段．约翰正打算安排他的N(1≤N≤25000)只奶牛来值班,打扫 打扫牛棚卫生．每只奶牛都有自己的空闲时间段[Si,Ei](1≤Si≤Ei≤T),只能把空闲的奶牛安排出来值班．而且,每个时间段必需有…

思路:可以贪心,也可以最短路. 贪心写法:因为在保证合法的前提下,我们选择的区间一定要右端点尽量靠后才行,于是我们每次就选择一个合法的并且右端点最靠后的区间就好了(如果没有合法的输出-1即可).时间复杂度O(nlogn)(排序是nlogn的,贪心是O(n)的). #include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> usi…

题目 1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚 Time Limit: 5 Sec  Memory Limit: 64 MB Description Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most…

Cleaning Shifts bzoj-3389 Usaco-2004Dec 题目大意:每天有n个时间段,每个时间段都必须安排一个奶牛值班.有m个奶牛,每个奶牛只有一个空闲时间s[i]~e[i],求至少动用多少奶牛. 注释:$1\le n\le 10^6$,$1\le m\le 25,000$. 想法:神题 我们将所有的时间点排成一排,然后对每一个i+1向i连一条无代价的边. 对于每一个s[i]向其对应的e[i]连边,代价为1. 然后求1到n的最短路即可 这样建图的道理:首先,从后面的时间点向…

设f[i]为i时刻最小花费 把牛按l升序排列,每头牛能用f[l[i]-1]+c[i]更新(l[i],r[i])的区间min,所以用线段树维护f,用排完序的每头牛来更新,最后查询E点即可 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N=10005; const long long inf=1e18; int n,st,ed; struct xd…

[BZOJ1672][Usaco2005 Dec]Cleaning Shifts Description Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire…

1672: [Usaco2005 Dec]Cleaning Shifts 清理牛棚 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 414  Solved: 191[Submit][Status] Description Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to…

BZOJ_1672_[Usaco2005 Dec]Cleaning Shifts 清理牛棚_动态规划+线段树 题意:  约翰的奶牛们从小娇生惯养,她们无法容忍牛棚里的任何脏东西．约翰发现,如果要使这群有洁癖的奶牛满意,他不得不雇佣她们中的一些来清扫牛棚, 约翰的奶牛中有N(1≤N≤10000)头愿意通过清扫牛棚来挣一些零花钱．由于在某个时段中奶牛们会在牛棚里随时随地地乱扔垃圾,自然地,她们要求在这段时间里,无论什么时候至少要有一头奶牛正在打扫．需要打扫的时段从某一天的第M秒开始,到第E秒结束f0…

P4644 [Usaco2005 Dec]Cleaning Shifts 清理牛棚 你有一段区间需要被覆盖(长度 <= 86,399) 现有 \(n \leq 10000\) 段小线段, 每段可以从 \(l_{i}\) 到 \(r_{i}\) 花费为 \(s_{i}\) 求覆盖整个区间的最小花费 错误日志: 初始化时应该是 \(dp[L - 1]\) 为 \(0\) 而不是 \(dp[1]\) 为 \(0\) , 因为只需要覆盖 \([L, R]\) Solution 设 \(dp[n]\) 表…

[Usaco2005 Dec] Cleaning Shifts 清理牛棚 题目描述 Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of th…

[Usaco2005 Dec]Cleaning Shifts 给出n段区间,左右端点分别为\(l_i,r_i\),以及选取这段区间的费用\(c_i\),现在要选出若干个区间,使其完全覆盖区间\([m,e]\),询问费用之和的最小值,\(1≤n≤10000,0≤m≤e≤86399\). 解 法一: 不妨把区间按左端点排序,如果在大区间范围外,可以筛除,虽然题目有保障,于是设\(f_i\)表示以第i个区间结尾,覆盖第i个区间前所有需要覆盖的位置的最少代价,于是有 \[f_i=\min_{j=1,r_…

题目 3390: [Usaco2004 Dec]Bad Cowtractors牛的报复 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 53  Solved: 37[Submit][Status] Description     奶牛贝茜被雇去建设N(2≤N≤1000)个牛棚间的互联网．她已经勘探出M(1≤M≤ 20000)条可建的线路,每条线路连接两个牛棚,而且会苞费C(1≤C≤100000)．农夫约翰吝啬得很,他希望建设费用最少甚至他都不想给贝茜…

因为是棵树 , 所以直接 dfs 就好了... ------------------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ; i <…

3390: [Usaco2004 Dec]Bad Cowtractors牛的报复 Time Limit: 1 Sec  Memory Limit: 128 MB Description     奶牛贝茜被雇去建设N(2≤N≤1000)个牛棚间的互联网．她已经勘探出M(1≤M≤ 20000)条可建的线路,每条线路连接两个牛棚,而且会苞费C(1≤C≤100000)．农夫约翰吝啬得很,他希望建设费用最少甚至他都不想给贝茜工钱． 贝茜得知工钱要告吹,决定报复．她打算选择建一些线路,把所有牛棚连接在一起,…

http://www.lydsy.com/JudgeOnline/problem.php?id=1672 dp很好想,但是是n^2的..但是可以水过..(5s啊..) 按左端点排序后 f[i]表示取第i个的最小费用 f[i]=min(f[j])+w[i] 当j的结束时间>=i的开始时间-1 答案就是所有的i满足i的结束时间>=结束时间-1 #include <cstdio> #include <cstring> #include <cmath> #inclu…

题目描述 Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn. Farmer John…

题目描述 Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn. Farmer John…

题目传送门 清理牛棚 题目描述 Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn.…

题目描述 Farmer John's cows, pampered since birth, have reached new heights of fastidiousness. They now require their barn to be immaculate. Farmer John, the most obliging of farmers, has no choice but hire some of the cows to clean the barn. Farmer John…

这很明显就是最大生成树= = CODE： #include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define maxn 1010#define maxm 20010struct edges{ int x,y,z;}edge[maxm];bool cmp(edges a,edges b){return (a.z>b.z);}in…

这道直接遍历一遍求出每个点的子节点数目就行了= = CODE： #include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define maxn 50010int b[maxn],q[maxn],id[maxn],ans[maxn];bool cmp(int x,int y){return q[x]<q[y];} int main…

[题目链接] http://www.lydsy.com/JudgeOnline/problem.php?id=3391 [题目大意] 给定一棵树,求分支size均不大于一半点数的点 [题解] 递归的同时计算各个分支size,如果出现大于一半点数的则给这个点打上标记 最后输出没有标记的点即可. [代码] #include <cstdio> #include <algorithm> #include <vector> using namespace std; const i…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=3391 题意: 给你一棵无根树,求分支size均不大于一半点数的点. 题解: 假定1为根. dfs时统计siz[i]和par[i]. 对于每个节点判断一下子树大小siz[son]和自己往上的子树大小n - siz[now]. AC Code: #include <iostream> #include <stdio.h> #include <string.h> #i…

裸的最大生成树,注意判不连通情况 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; const int N=20005; int n,m,f[N],con; long long ans; struct qwe { int u,v,w; }a[N]; bool cmp(const qwe &a,const qwe &b) { return a.w>…

链接 题意:给你一些区间,每个区间都有一个花费,求覆盖区间 \([S,T]\) 的最小花费 题解 先将区间排序 设 \(f[i]\) 表示决策到第 \(i\) 个区间,覆盖满 \(S\dots R[i]\) 的最小花费 显然 \(f[i]=\min_{R[j]\ge L[i]}f[j]+w[i]\) 按照区间建线段树,插右端点即可 #include<bits/stdc++.h> #define REP(i,a,b) for(int i(a);i<=(b);++i) using names…