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…

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…

POJ 3045 Cow Acrobats 这是个贪心的题目,和网上的很多题解略有不同,我的贪心是从最下层开始,每次找到能使该层的牛的风险最小的方案, 记录风险值,上移一层,继续贪心. 最后从遍历每一层的风险值,找到其中的最大值 我一开始对sum-p[i].a-p[i].b从小到大排序,这样第一次取出的就是能使最下层的牛的风险最小的方案,在上移一层时,这一层的风险值   为sum-p[i].a-p[i].b-p[0].a,由于p[0].a是固定值,所以第二次直接取出的就是能使该层的牛的风险最小的…

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…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8122   Accepted: 3674 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…

POJ 1847 Tram (最短路径) Description Tram network in Zagreb consists of a number of intersections and rails connecting some of them. In every intersection there is a switch pointing to the one of the rails going out of the intersection. When the tram ent…

POJ 3660 Cow Contest / HUST 1037 Cow Contest / HRBUST 1018 Cow Contest(图论,传递闭包) Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has…

POJ 3259 Wormholes(最短路径,求负环) Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE…

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(…

POJ 3176 Cow Bowling 题目简化即为从一个三角形数列的顶端沿对角线走到底端,所取得的和最大值 7 * 3 8 * 8 1 0 * 2 7 4 4 * 4 5 2 6 5 该走法即为最大值 分析:简单的动态规划,从上往下一层一层的考虑,对于每一行的最左边和最右边只有一种走法,只需要简单的相加, 对于中间的数要考虑是加上左上角的数还是加右上角的数,加上两者中的较大者 代码: #include<iostream> #include<cstdio> #include<…

POJ-2184 [题意]: 有n头牛,每头牛有自己的聪明值和幽默值,选出几头牛使得选出牛的聪明值总和大于0.幽默值总和大于0,求聪明值和幽默值总和相加最大为多少. [分析]:变种的01背包,可以把幽默度看成体积,智商看成价值,那么就转换成求体积和价值都为正值的最大值的01背包了. 以 TS 作为体积,TF作为价值,在保证体积.价值非负的情况下,求解 sum,取其所有情况的最大值. 难点: 1)体积出现负数,将区间改变 [-100000, 100000] ---> [0, 200000]. (注…

POJ 2375 Cow Ski Area id=2375" target="_blank" style="">题目链接 题意:给定一个滑雪场,每一个点能向周围4个点高度小于等于这个点的点滑,如今要建电缆,使得随意两点都有路径互相可达,问最少须要几条电缆 思路:强连通缩点.每一个点就是一个点.能走的建边.缩点后找入度出度为0的个数的最大值就是答案.注意一開始就强连通了答案应该是0 代码: #include <cstdio> #includ…

Poj 3613 Cow Relays (图论) 题目大意 给出一个无向图,T条边,给出N,S,E,求S到E经过N条边的最短路径长度 理论上讲就是给了有n条边限制的最短路 solution 最一开始想到是的去直接统计最短路经过了多少条边,结果,,, 还是太年轻了... 不过,看数据范围只有1000,那么floyd是首选 回顾Floyd算法流程,其中的i到j松弛操作是通过k完成的 那么松弛一次就利用一个k点,我现在要经过n条边,那么松弛n次即可 详细说就是更新一次之后,把f[i][j]拷贝到原来的…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5989   Accepted: 3234 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others…

http://poj.org/problem?id=3613 题目大意: 有T条路.从s到e走n步,求最短路径. 思路: 看了别人的... 先看一下Floyd的核心思想: edge[i][j]=min(edge[i][j],edge[i][k]+edge[k][j])  i到j的最短路是i到j的直接路径或者经过k点的间接路径.可是矩阵的更新总是受到上一次更新的影响 假设每次的更新都存进新矩阵,那么edge[i][k]+edge[k][j]是不是表示仅仅经过三个点两条边的路径呢? min(edge…

Silver Cow Party Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 1   Accepted Submission(s) : 1 Problem Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is goin…

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…

题目链接:http://poj.org/problem?id=3615 题意:大致题意:有N个木桩,和M个木桩对之间的高度差(从x跳到y需要往上跳的高度).从x跳跃到y的路径消耗的体力值是路径中的一个最大高度差.求一条消耗体力最小的路径. #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; #define INF 0x3f3f3f3f ][]; int main…

题目链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10066   Accepted: 5682 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kn…

题目连接 http://poj.org/problem?id=1985 Cow Marathon Description After hearing about the epidemic of obesity in the USA, Farmer John wants his cows to get more exercise, so he has committed to create a bovine marathon for his cows to run. The marathon ro…

题目链接:http://poj.org/problem?id=3615 思路:map[i][j]表示顶点i,j之间的最高的障碍物,于是题目要求的是最高障碍物的最小值,不就是min(map[i][j],max(map[i][k],map[k][j]))嘛. http://paste.ubuntu.com/5925170/…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5989   Accepted: 3234 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others…

题目链接:http://poj.org/problem?id=2184 Cow Exhibition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9479   Accepted: 3653 Description "Fat and docile, big and dumb, they look so stupid, they aren't much  fun..."  - Cows with Guns by…

Wormholes Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 29971   Accepted: 10844 Description While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way p…

Cow Multiplication Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13312   Accepted: 9307 Description Bessie is tired of multiplying pairs of numbers the usual way, so she invented her own style of multiplication. In her style, A*B is eq…

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…

Description The cow bicycling team consists of N (1 <= N <= 20) cyclists. They wish to determine a race strategy which will get one of them across the finish line as fast as possible. Like everyone else, cows race bicycles in packs because that's th…

http://poj.org/problem?id=3613 题意: 求经过k条路径的最短路径. 思路: 如果看过<矩阵乘法在信息学的应用>这篇论文就会知道 现在我们在邻接矩阵中保存距离,那么按照上面计算,不就是k路径的最短路径了吗? 每次用folyd去最小值,至于k次就是相乘,用快速幂. #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #incl…

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…