POJ-3186_Treats for the Cows】的更多相关文章

POJ 2387 Til the Cows Come Home (图论,最短路径) Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to g…
POJ.2387 Til the Cows Come Home (SPFA) 题意分析 首先给出T和N,T代表边的数量,N代表图中点的数量 图中边是双向边,并不清楚是否有重边,我按有重边写的. 直接跑spfa,dij,floyd都可以. 求1到N的最短路. 代码总览 #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <stack>…
题目链接:http://poj.org/problem?id=2387 Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 45414   Accepted: 15405 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible…
题目连接: http://poj.org/problem?id=2387 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get ba…
传送门 Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 46727   Accepted: 15899 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for…
Til the Cows Come Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 33015   Accepted: 11174 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the…
Til the Cows Come Home 题目链接: http://acm.hust.edu.cn/vjudge/contest/66569#problem/A Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie…
Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer Joh…
Til the Cows Come Home Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly…
Aggressive cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20485   Accepted: 9719 Description Farmer John has built a new long barn, with N (2 <= N <= 100,000) stalls. The stalls are located along a straight line at positions x1,..…
Description:     就是给你一个数,你可以把它自乘,也可以把他乘或除以任意一个造出过的数,问你最多经过多少次操作能变换成目标数 思路:这题真的不怎么会啊.n = 20000,每一层都有很多个扩展状态,裸宽搜会被T,启发式函数又设计不出来…… 看了一个Vjudge上的代码才知道这题怎么写. 就是每一个状态是由最多两个数转化而来的,所以可以把两个数看做一个状态. 用一个多元组$node(x,y,g,h)$表示状态,$x, y$分别表示两个数中的较大数和较小数,然后$g$表示转换成当前的…
题目链接:http://poj.org/problem?id=3186 题目大意:给出的一系列的数字,可以看成一个双向队列,每次只能从队首或者队尾出队,第n个出队就拿这个数乘以n,最后将和加起来,求最大和. 解题思路:有两种写法: ①这是我一开始想的,从外推到内,设立数组dp[i][j]表示剩下i~j时的最优解,则有状态转移方程: dp[i][j]=dp[i][j]=max(dp[i-1][j]+a[i-1]*(n-(j-i+1)),dp[i][j+1]+a[j+1]*(n-(j+1-i)))…
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer John's field ha…
题目链接:http://poj.org/problem?id=2387 题目大意:起点一定是1,终点给出,然后求出1到所给点的最短路径. 注意的是先输入边,在输入的顶点数,不要弄反哦~~~ #include <iostream> #include <cstdio> using namespace std; ][],Min,node[],vis[],t,q; ; void set() { ; i<=; i++) { node[i]=INF; vis[i]=; ; j<=;…
题目链接:http://poj.org/problem?id=2387 题目大意:给你t条边(无向图),n个顶点,让你求点1到点n的最短距离. 解题思路:裸的dijsktra,注意判重边. 代码: #include<cstdio> #include<cmath> #include<cctype> #include<cstring> #include<iostream> #include<algorithm> #include<v…
http://poj.org/problem?id=2186 tarjan求强连通分量. 因为SD省选用WinXP+Cena评测而且不开栈,所以dfs只好写手动栈了. 写手动栈时思路清晰一点应该是不会出错的吧... 这里tarjan要开两个栈,一个是tarjan用来记录强连通分量的栈,另一个是记录dfs路径的栈. cur记录当前弧,fa记录dfs树上的father. 对于扫到一个环不需要dfs下去的情况就直接处理,对于需要dfs下去的情况就把下一个点压入栈,记录这下一个点的fa为当前点,同时当前…
Corral the Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1352   Accepted: 565 Description Farmer John wishes to build a corral for his cows. Being finicky beasts, they demand that the corral be square and that the corral contain a…
Time Limit: 1000MSMemory Limit: 65536K Total Submissions: 59755Accepted: 20336 Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie need…
Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.…
[题目链接] http://poj.org/problem?id=2182 [算法] 树状数组 + 二分 [代码] #include <algorithm> #include <bitset> #include <cctype> #include <cerrno> #include <clocale> #include <cmath> #include <complex> #include <cstdio> #…
传送门:http://poj.org/problem?id=2387 题目大意: 给定无向图,要求输出从点n到点1的最短路径. 注意有重边,要取最小的. 水题..对于无向图,从1到n和n到1是一样的. 直接Dijkstra即可 #include<cstdio> #include<cstring> const int MAXN=1000+10; const int INF=999999; int map[MAXN][MAXN]; int dis[MAXN]; int main() {…
http://poj.org/problem?id=3621 题意:有n个点m条有向边,每个点有一个点权val[i],边有边权w(i, j).找一个环使得Σ(val) / Σ(w)最大,并输出. 思路:和之前的最优比率生成树类似,还是构造成这样的式子:F(L) = Σ(val[i] * x[i]) - Σ(w[i] * x[i] * L) = Σ(d[i]) * x[i] (d[i] = val[i] - w[i] * L),要使得L越大越好. 那么当L越大的时候,F(L)就越小,如果F(L)大…
Aggressive cows 直接上中文了 Descriptions 农夫 John 建造了一座很长的畜栏,它包括N (2 <= N <= 100,000)个隔间,这些小隔间依次编号为x1,...,xN (0 <= xi <= 1,000,000,000). 但是,John的X (2 <= X <= N)头牛们并不喜欢这种布局,而且几头牛放在一个隔间里,他们就要发生争斗.为了不让牛互相伤害.John决定自己给牛分配隔间,使任意两头牛之间的最小距离尽可能的大,那么,这个…
Description Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer Joh…
传送门 题意: 农场主 FJ 有 n 头奶牛,现在给你 m 对关系(x,y)表示奶牛x的产奶速率高于奶牛y: FJ 想按照奶牛的产奶速率由高到低排列这些奶牛,但是这 m 对关系可能不能精确确定这 n 头奶牛的关系: 问最少需要额外增加多少对关系使得可以确定这 n 头奶牛的顺序: 题解: 之所以做这道题,是因为在补CF的题时用到了bitset<>: 搜这个容器的用法是看到了一篇标题为POJ-3275:奶牛排序Ranking the Cows(Floyd.bitset)的文章: 正好拿着道题练练b…
题目链接:http://poj.org/problem?id=2387 题意:有n个城市点,m条边,求n到1的最短路径.n<=1000; m<=2000 就是一个标准的最短路模板. #include<iostream> #include<algorithm> #include<cstring> #include<vector> #include<queue> using namespace std; ; const int INF =…
Lost Cows Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10544   Accepted: 6754 Description N (2 <= N <= 8,000) cows have unique brands in the range 1..N. In a spectacular display of poor judgment, they visited the neighborhood 'wateri…
Popular Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 25945   Accepted: 10612 Description Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M &…
Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible. Farmer John's field ha…
原题链接:Til the Cows Come Home 题目大意:有  个点,给出从  点到  点的距离并且  和  是互相可以抵达的,问从  到  的最短距离. 题目分析:这是一道典型的最短路径模版题,需要注意的是:使用dijkstra算法求解需要考虑有重复边问题,而使用bellman-ford算法 和 spfa算法 可以忽略这个问题. 代码如下: // Dijkstra #include <iostream> using namespace std; const int INTFY = 1…