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…

Cow Ski Area Time Limit: 1000ms Memory Limit: 65536KB This problem will be judged on PKU. Original ID: 237564-bit integer IO format: %lld      Java class name: Main   Farmer John's cousin, Farmer Ron, who lives in the mountains of Colorado, has recen…

题目地址:POJ 2375 对每一个点向与之相邻并h小于该点的点加有向边. 然后强连通缩点.问题就转化成了最少加几条边使得图为强连通图,取入度为0和出度为0的点数的较大者就可以.注意,当强连通分量仅仅有一个的时候.答案是0,而不是1. 代码例如以下: #include <iostream> #include <string.h> #include <math.h> #include <queue> #include <algorithm> #in…

题目链接:http://poj.org/problem?id=2375题目大意:一片滑雪场,奶牛只能向相邻的并且不高于他当前高度的地方走.想加上缆车是的奶牛能从低的地方走向高的地方,求最少加的缆车数,是的奶牛可以从任意一个角落到达任意另外的角落解题思路:奶牛可以向相邻的不高于他的地方走,相当于u,v之前连通.若想加上缆车是图成为连通图,数目就是max(root, leave);思路同POJ 1236 代码如下: #include<stdio.h> #include<string.h>…

这个题目用tarjan找联通块,缩点,然后统计出入度为0的点理论上是可行的,但问题是会暴栈.考虑到这个题目的特殊性,可以直接用一次bfs找到数字相同且联通的块,这就是一个联通块,然后缩点,统计出入度即可. #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; const int maxn=1e3+9; int a[maxn]…

题目大意:一个W*L的山,每个山有个高度,当且仅当一个山不比它相邻(有公共边的格子)的山矮时能够滑过去,现在可以装化学电梯来无视山的高度滑雪,问最少装多少电梯使得任意两点都可到达 思路:最后一句话已经把强连通模型裸裸地说出来了 那问题变成了一个图最小加几条边变成强连通图的经典问题,比较一下出度为0和入度为0的点的个数的大小即可,还有个特例只有一个SCC的情况 #include<cstdio> #include<string.h> #include<iostream> #…

                                    Cow Ski Area Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3323   Accepted: 919 Description Farmer John's cousin, Farmer Ron, who lives in the mountains of Colorado, has recently taught his cows to s…

Description Farmer John's cousin, Farmer Ron, who lives in the mountains of Colorado, has recently taught his cows to ski. Unfortunately, his cows are somewhat timid and are afraid to ski among crowds of people at the local resorts, so FR has decided…

http://poj.org/problem?id=2375 题意:一个500*500的矩形,每个格子都有一个高度,不能从高度低的格子滑到高度高的格子(但相等高度可以滑),已知可以在2个相邻格子上加桥,使得无视他们的高度就可以互相滑,问最少加多少桥可以使得在任一个格子上都能到达任一个格子. 分析:很容易看出这就是相当于在一个有向图上至少加多少边可以使得其强联通,ans=max(入度0的点数,出度为0的点数),很好理解,可以把出度为0的点挂一条边到入度为0的点上,多了的随便挂.那么现在面临的问题的…

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…

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 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 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 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…

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…

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…

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…

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…

Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3660 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some co…

原题链接:http://poj.org/problem?id=3660 Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 8395   Accepted: 4734 Description N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all kno…

Cow Contest POJ - 3660 :http://poj.org/problem?id=3660   参考:https://www.cnblogs.com/kuangbin/p/3140837.html   题意: n头牛,有m对牛进行了比赛,现在告诉你每队牛比赛的结果,A胜B,问有几头牛的排名可以确定. 思路: 题目给出了m对的相对关系,求有多少个排名是确定的. 使用floyed求一下传递闭包.如果这个点和其余的关系都是确定的,那么这个点的排名就是确定的. #include <al…

Cow Bowling Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13016   Accepted: 8598 Description The cows don't use actual bowling balls when they go bowling. They each take a number (in the range 0..99), though, and line up in a standard…

链接 Cow Contest Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu 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. Eac…

题目链接:http://poj.org/problem?id=1985 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 route will include a pair o…

链接:poj 3660 题意:给定n头牛,以及某些牛之间的强弱关系.按强弱排序.求能确定名次的牛的数量 思路:对于某头牛,若比它强和比它弱的牛的数量为 n-1,则他的名次能够确定 #include<stdio.h> #include<limits.h> int a[110][110]; int main() { int n,m,i,j,k,s,sum; while(scanf("%d%d",&n,&m)!=EOF){ for(i=1;i<=…

Cow Contest Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7690   Accepted: 4288 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…

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19226   Accepted: 8775 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of …