题目链接:http://poj.org/problem?id=3268 题意 :有N头奶牛,M条单向路.X奶牛开party,其他奶牛要去它那里.每头奶牛去完X那里还要返回.去回都是走的最短路.现在问这里面哪头奶牛走的路最长. 题解:对每个奶牛i与X做两次spfa.去回各一次.然后统计最长的..板子稍微改一改//但是我还是T了好几发,因为初始化数组的时候maxn开大了..QAQ.改小了就过了. 代码: #include<iostream> #include<stack> #inclu…

Description One cow ≤ N ≤ ) conveniently numbered ..N ≤ X ≤ N). A total of M ( ≤ M ≤ ,) unidirectional (one-way roads connects pairs of farms; road i requires Ti ( ≤ Ti ≤ ) units of time to traverse. Each cow must walk to the party and, when the part…

题目:http://poj.org/problem?id=3268 题解:使用 priority_queue队列对dijkstra算法进行优化 #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <functional> #include <cstring> using namespace std; + ; ; str…

题目链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19211   Accepted: 8765 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow…

POJ 3268 Silver Cow Party (最短路径) 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 M (1 ≤ M ≤ 100,000) unidirectional (one-way roads c…

POJ 3268 Silver Cow Party 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 M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects…

原题链接:http://poj.org/problem?id=3268 Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 15545   Accepted: 7053 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow…

Description Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example: Georgia an…

Description 有两堆石子,数量任意,可以不同.游戏开始由两个人轮流取石子.游戏规定,每次有两种不同的取法,一是可以在任意的一堆中取走任意多的石子:二是可以在两堆中同时取走相同数量的石子.最后把石子全部取完者为胜者.现在给出初始的两堆石子的数目,如果轮到你先取,假设双方都采取最好的策略,问最后你是胜者还是败者. Input 输入包含若干行,表示若干种石子的初始情况,其中每一行包含两个非负整数a和b,表示两堆石子的数目,a和b都不大于1,000,000,000. Output 输出对应也有…

[BZOJ3939][Usaco2015 Feb]Cow Hopscotch Description Just like humans enjoy playing the game of Hopscotch, Farmer John's cows have invented a variant of the game for themselves to play. Being played by clumsy animals weighing nearly a ton, Cow Hopscotc…

[BZOJ1604][Usaco2008 Open]Cow Neighborhoods 奶牛的邻居 Description 了解奶牛们的人都知道,奶牛喜欢成群结队．观察约翰的N(1≤N≤100000)只奶牛,你会发现她们已经结成了几个“群”．每只奶牛在吃草的时候有一个独一无二的位置坐标Xi,Yi(l≤Xi,Yi≤[1．.10^9]:Xi,Yi∈整数．当满足下列两个条件之一,两只奶牛i和j是属于同一个群的:   1．两只奶牛的曼哈顿距离不超过C(1≤C≤10^9),即lXi - xil+IYi -…

Silver Cow Party Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Submit Status Practice POJ 3268 Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to b…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13611   Accepted: 6138 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…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12674   Accepted: 5651 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 …

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13982   Accepted: 6307 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 …

                                                                                                   Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 19325   Accepted: 8825 Description One cow from each of N farms (1 ≤ N ≤…

Silver Cow Party 题目链接: http://acm.hust.edu.cn/vjudge/contest/122685#problem/D 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 M (1 ≤…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17017   Accepted: 7767 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 …

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions:28457   Accepted: 12928 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 …

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…

Silver Cow Party 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 M (1 ≤ M ≤ 100,000) unidirectional (one-way roads connects pairs of farms; road…

Silver Cow Party Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 22864   Accepted: 10449 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…

Silver cow party 迪杰斯特拉+反向 题意 有n个农场,编号1到n,每个农场都有一头牛.他们想要举行一个party,其他牛到要一个定好的农场中去.每个农场之间有路相连,但是这个路是单向的,并且去了还得回来,求花费时间最多是多少? 解题思路 很容易想明白需要分两步 第一步:算出目的点到其他点的最短距离, 这一步很好实现,直接使用Dijkstra即可 第二部:算出其他点到终点的最短距离. 关键就在第二部.迪杰斯特拉算的是某一点到其他所有点的最短距离,而这次我们是求的其他点到某一点的最短…

DP/单调队列优化 首先不考虑奶牛的喜欢区间,dp方程当然是比较显然的:$ f[i]=min(f[k])+1,i-2*b \leq k \leq i-2*a $  当然这里的$i$和$k$都是偶数啦~这个应该很好理解吧……每次喷灌的都是一个偶数长度的区间嘛…… 那么加上奶牛的喜欢区间的话,只需这样:当$ i>cow[j].x $时,令$ i=cow[j].y , j++$ 也就是说中间的位置全部不考虑放喷灌器. 显然我们对于每个节点的 k 是可以用单调队列维护的!嗯看到这里的同学可以先自己试着去…

可持久化线段树 可持久化线段树是一种神奇的数据结构,它跟我们原来常用的线段树不同,它每次更新是不更改原来数据的,而是新开节点,维护它的历史版本,实现“可持久化”.(当然视情况也会有需要修改的时候) 可持久化线段树的应用有很多,仅以区间第K大这种简单的问题来介绍这种数据结构. 我们原本建立的线段树是表示区间的,或者说,维护的是[位置],存的是每个位置上的各种信息.它的优点是满足区间加法,但不满足区间减法,所以我们这里要换一种建树方式:对于每个区间[1,i]建立一棵权值线段树.这个线段树的作用其实就…

[算法]高斯消元 [题解] 高斯消元经典题型:异或方程组 poj 1222 高斯消元详解 异或相当于相加后mod2 异或方程组就是把加减消元全部改为异或. 异或性质:00 11为假,01 10为真.与1异或取反,与0异或不变. 建图:对于图上每个点x列一条异或方程,未知数为n个灯按不按,系数为灯i按了点x变不变,该行结果n+1为初始状态.(所以a[x][y]其实表示x和y是否存在异或关系) 建图原理见上面链接. 寻找:因题目保证有解,而系数只有0或1,所以不用找最大,找到一个非0系数即可. 消元…

[算法]平衡树(treap) [题解]treap知识见数据结构 在POJ把语言从G++换成C++就过了……??? #include<cstdio> #include<algorithm> #include<ctime> using namespace std; ; ]; int n,m,root,sz,s,tt,d[maxn]; void rr(int &tt)//右旋 { int k=t[tt].l; t[tt].l=t[k].r; t[k].r=tt; tt…

网络流/最大流 愚人节快乐XD 这题是给一个混合图(既有有向边又有无向边),让你判断是否有欧拉回路…… 我们知道如果一个[连通]图中每个节点都满足[入度=出度]那么就一定有欧拉回路…… 那么每条边都可以贡献一个出度出来,对于一条边u->v: 连S->edge cap=1; 如果是有向边,就连 edge->v cap=1; 否则(无向边)连edge->u cap=1, edge->v cap=1; 然后每个点的总度数我们是知道的……那么它最后的[出度]就等于 总度数/2.(这个…

[题目] ExponentiationTime Limit: 500MS Memory Limit: 10000KTotal Submissions: 123707 Accepted: 30202Description Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the nat…

[算法]矩阵快速幂 [题解] 根据f[n]=f[n-1]+f[n-2],可以构造递推矩阵: $$\begin{vmatrix}1 & 1\\ 1 & 0\end{vmatrix} \times \begin{vmatrix}f_n \\ f_{n-1} \end{vmatrix}=\begin{vmatrix}f_{n+1}\\f_n\end{vmatrix}\\$$ 写成幂形式: $$\begin{vmatrix}1 & 1\\ 1 & 0\end{vmatrix}^n…