题目描述 Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hu…

P2888 [USACO07NOV]牛栏Cow Hurdles Floyd $n<=300$?果断Floyd 给出核心式,自行体会 $d[i][j]=min(d[i][j],max(d[i][k],d[k][j]))$ #include<iostream> #include<cstdio> #include<cstring> using namespace std; int max(int a,int b){return a>b?a:b;} int min(…

题目戳 题目描述 Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over th…

题目描述 Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over the hu…

OJ题号:洛谷2888 思路:修改Floyd,把边权和改为边权最大值.另外注意是有向图. #include<cstdio> #include<algorithm> using namespace std; #define inf 0x7fffffff int main() { int n,m,t; scanf("%d%d%d",&n,&m,&t); ][n+]; ;i<=n;i++) { ;j<=n;j++) { d[i][j…

Luogu P2889 [USACO07NOV]挤奶的时间Milking Time 题目描述 传送门Bessie is such a hard-working cow. In fact, she is so focused on maximizing her productivity that she decides to schedule her next N (1 ≤ N ≤ 1,000,000) hours (conveniently labeled 0..N-1) so that she…

直接floyd.. ---------------------------------------------------------------------------- #include<cstdio> #include<cstring> #include<algorithm> #include<iostream>   #define rep( i , n ) for( int i = 0 ;  i < n ; ++i ) #define clr(…

1641: [Usaco2007 Nov]Cow Hurdles 奶牛跨栏 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 424  Solved: 272[Submit][Status] Description Farmer John 想让她的奶牛准备郡级跳跃比赛,贝茜和她的伙伴们正在练习跨栏.她们很累,所以她们想消耗最少的能量来跨栏. 显然,对于一头奶牛跳过几个矮栏是很容易的,但是高栏却很难.于是,奶牛们总是关心路径上最高的栏的高度. 奶牛的训练场…

Cow Hurdles Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9337   Accepted: 4058 Description Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are…

https://www.luogu.org/problemnew/show/P2886 给定无向连通图,求经过k条边,s到t的最短路 Floyd形式的矩阵乘法,同样满足结合律,所以可以进行快速幂. 离散化大可不必sort+unique,因为我们甚至并不在乎相对大小,只是要一个唯一编号,可以开一个桶记录 之所以进行k-1次快速幂,是因为邻接矩阵本身就走了一步了. #include<iostream> #include<cstring> #include<cstdio> u…

BZOJ 1706权限题. 倍增$floyd$. 首先这道题有用的点最多只有$200$个,先离散化. 设$f_{p, i, j}$表示经过$2^p$条边从$i$到$j$的最短路,那么有转移$f_{p, i, j} = min(f_{p - 1, i, k} + f_{p - 1, k, j})$. 然后做一个类似于快速幂的东西把$n$二进制拆分然后把当前的$f$代进去转移. 可以设一个$g_{i, j}$表示当前从$i$到$j$的最短路,为了保证转移顺序的正确,可以把$g$抄出来到$h$中,然后…

Description Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be able to use as little energy as possible to jump over…

题面:P5200 [USACO19JAN]Sleepy Cow Sorting 题解: 最小操作次数(记为k)即为将序列倒着找第一个P[i]>P[i+1]的下标,然后将序列分成三部分:前缀部分(待转移部分),k,后缀部分(不需转移部分). 用树状数组维护前缀部分每一个数挪到后缀部分所需的最小代价(即插到第一个小于它的数前)(这部分完全可以用线段树做). 代码: #include<cstdio> #include<cstring> #include<iostream>…

https://www.luogu.org/problemnew/show/P2879 差分 | 线段树 #include <iostream> #include <cstdio> #include <algorithm> using namespace std; ; #define gc getchar() struct Node {int l, r;}A[N]; int n, my, Maxh, R; int H[N]; inline int read() { ;…

http://poj.org/problem?id=3615 (题目链接) 题意 给出一张有向图,求从u到v最大边最小的路径的最大边.→_→不会说话了.. Solution 好久没写Floyd了,水一发.邻接表都不用打... 代码 // poj3615 #include<algorithm> #include<iostream> #include<cstdlib> #include<cstring> #include<cstdio> #inclu…

Description Farmer John 想让她的奶牛准备郡级跳跃比赛,贝茜和她的伙伴们正在练习跨栏.她们很累,所以她们想消耗最少的能量来跨栏. 显然,对于一头奶牛跳过几个矮栏是很容易的,但是高栏却很难.于是,奶牛们总是关心路径上最高的栏的高度. 奶牛的训练场中有 N (1 ≤ N ≤ 300) 个站台,分别标记为1..N.所有站台之间有M (1 ≤ M ≤ 25,000)条单向路径,第i条路经是从站台Si开始,到站台Ei,其中最高的栏的高度为Hi (1 ≤ Hi ≤ 1,000,000)…

Description Farmer John 想让她的奶牛准备郡级跳跃比赛,贝茜和她的伙伴们正在练习跨栏.她们很累,所以她们想消耗最少的能量来跨栏. 显然,对于一头奶牛跳过几个矮栏是很容易的,但是高栏却很难.于是,奶牛们总是关心路径上最高的栏的高度. 奶牛的训练场中有 N (1 ≤ N ≤ 300) 个站台,分别标记为1..N.所有站台之间有M (1 ≤ M ≤ 25,000)条单向路径,第i条路经是从站台Si开始,到站台Ei,其中最高的栏的高度为Hi (1 ≤ Hi ≤ 1,000,000)…

http://www.lydsy.com/JudgeOnline/problem.php?id=1641 这种水题无意义... #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm> #include <queue> using namespace std…

NOIP2013货车运输.仅仅只是数据范围小了很多. 不到150s打完而且AC. . 额.当然.我写的是Floyd. 写LCA的真过分. #include <cstdio> #include <cstring> #include <algorithm> #define N 305 #define inf 0x3f3f3f3f using namespace std; int n,m,q; int map[N][N]; int main() { // freopen(&q…

题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1641 题意: 给你一个有向图,n个点(n <= 300),m条边,边权为h[i]. t个询问(a,b).让你找一条从a到b的路径,使路径上最大的边最小,输出这个最小值. 题解: 新版floyd. dis[a][b]表示从a到b的路径中,最大边的最小值. 分别枚举k,i,j,然后取最小: dis[i][j] = min(dis[i][j], max(dis[i][k],dis[k][j])…

http://poj.org/problem?id=3615 floyd 最短路径的变形 dist[i][j]变化为 : i j之间的最大边 那么输入的时候可以直接把dist[i][j] 当作i j 之间的边进行输入 转移方程 dist[i][j] = max(dist[i][j], min(dist[i][k], dist[k][j])) #include <iostream> #include <string.h> #include <stdio.h> #defin…

先跑一遍取max版的Floyd,直接用数组回答询问即可 #include<iostream> #include<cstdio> using namespace std; const int N=305,inf=1e9; int n,m,q,a[N][N]; int read() { int r=0,f=1; char p=getchar(); while(p>'9'||p<'0') { if(p=='-') f=-1; p=getchar(); } while(p>…

[题解] 弗洛伊德.更新距离的时候把$f[i][j]=min(f[i][j],f[i][k]+f[k][j])$改为$f[i][j]=min(f[i][j],max(f[i][k],f[k][j]))$. #include<cstdio> #include<algorithm> #include<cstring> #define N (400) #define rg register using namespace std; int n,m,t,a[N][N]; inl…

Cow Hurdles Time Limit: 1000MS Memory Limit: 65536K Description Farmer John wants the cows to prepare for the county jumping competition, so Bessie and the gang are practicing jumping over hurdles. They are getting tired, though, so they want to be a…

转自:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★2544 最短路 基础最短路★3790 最短路径问题基础最短路★2066 一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★2112 HDU Today基础最短路★1874 畅通工程续基础最短路★1217 Arbitrage 货币交换 Floyd (或者 Bellman-Ford 判环)★124…

转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并查集======================================[HDU]1213   How Many Tables   基础并查集★1272   小希的迷宫   基础并查集★1325&&poj1308  Is It A Tree?   基础并查集★1856   More i…

大神们都在刷usaco,我也来水一水 1606: [Usaco2008 Dec]Hay For Sale 购买干草   裸背包 1607: [Usaco2008 Dec]Patting Heads 轻拍牛头 神转化,筛法 1609: [Usaco2008 Feb]Eating Together麻烦的聚餐  LIS 1610: [Usaco2008 Feb]Line连线游戏 排序 1611: [Usaco2008 Feb]Meteor Shower流星雨  BFS 1612: [Usaco2008…

先简单介绍一下最短路径: 最短路径是啥?就是一个带边值的图中从某一个顶点到另外一个顶点的最短路径. 官方定义:对于内网图而言,最短路径是指两顶点之间经过的边上权值之和最小的路径. 并且我们称路径上的第一个顶点为源点,最后一个顶点为终点. 由于非内网图没有边上的权值,所谓的最短路径其实是指两顶点之间经过的边数最少的路径. 我们时常会面临着对路径选择的决策问题,例如在中国的一些一线城市如北京.上海.广州.深圳等,一般从A点到到达B点都要通过几次地铁.公交的换乘才可以到达. 有些朋友想用最短对的时间,…

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…

=============================以下是最小生成树+并查集====================================== [HDU] How Many Tables 基础并查集★ 小希的迷宫 基础并查集★ &&poj1308 Is It A Tree? 基础并查集★ More is better 基础并查集★ Constructing Roads 基础最小生成树★ 畅通工程 基础并查集★ 还是畅通工程 基础最小生成树★ 畅通工程 基础最小生成树★ 畅通…