https://www.luogu.org/problem/show?pid=2863#sub 题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform th…

传送门 题目大意:形成一个环的牛可以跳舞,几个环连在一起是个小组,求几个小组. 题解:tarjian缩点后,求缩的点包含的原来的点数大于1的个数. 代码: #include<iostream> #include<cstdio> #include<cstring> #define maxn 10009 using namespace std; int n,m,sumedge,top,sumclr,tim,ans; int Stack[maxn],instack[maxn]…

代码是粘的,庆幸我还能看懂. #include<iostream> #include<cstdio> #include<cmath> #include<algorithm> #include<stack> #include<vector> using namespace std; struct node { int num,par; }p[]; ,x,y,cnt; vector<]; ]; ]; stack<int>…

每日一题 day11 打卡 Analysis 好久没大Tarjan了,练习练习模板. 只要在Tarjan后扫一遍si数组看是否大于1就好了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 10000+10 #define maxm 50000+10 using namespace std; inline int read() { ;…

一道tarjan的模板水题 在这里还是着重解释一下tarjan的代码 #include<iostream> #include<cstdio> #include<algorithm> using namespace std; int n,m; int cnt;//记录强联通分量的个数 int visitnum;//遍历的步数 int dfn[100010];//记录元素第一次被访问的步数 int low[100010];//包含i的强联通分量最早被访问的步数 int nu…

本来分好组之后,就确定好了每个人要学什么,我去学数据结构啊. 因为前一段时间遇到一道题是用Lca写的,不会,就去学. 然后发现Lca分为在线算法和离线算法,在线算法有含RMQ的ST算法,前面的博客也写了.离线算法是基于DFS的Tarjan算法. 然后就打算去学一下Tarjan,因为以前也看过但是没看完,就打算学一下,因为Tarjan算法是图论的内容,然后就让图论选手教了我一下大环套小环的怎么推,然后就尴尬了. 我是学数据结构的,没有要去抢图论的内容学... 我也看了线段树了啊. 学完这个Tarj…

洛谷——P2863 [USACO06JAN]牛的舞会The Cow Prom 题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round D…

P2863 [USACO06JAN]牛的舞会The Cow Prom 求点数$>1$的强连通分量数,裸的Tanjan模板. #include<iostream> #include<cstdio> #include<cstring> using namespace std; int min(int &a,int &b){return a<b?a:b;} #define N 10002 #define M 50002 int n,m,clo,df…

题目链接:https://www.luogu.org/problemnew/show/P2863 求强连通分量大小>自己单个点的 #include <stack> #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const int maxn = 100000 + 10; struct edge…

题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round Dance. Only cows can perform the Round D…

题目传送门 这个题还是个缩点的板子题...... 答案就是size大于1的强连通分量的个数 加一个size来统计就好了 #include <iostream> #include <cstdlib> #include <cstdio> using namespace std; const int N=1e5+5; const int M=5e5+5; struct edge{ int to,next; }e[M]; int n,m,dfn[N],low[N],cnt,he…

P2863 [USACO06JAN]牛的舞会The Cow Prom 123通过 221提交 题目提供者 洛谷OnlineJudge 标签 USACO 2006 云端 难度 普及+/提高 时空限制 1s / 128MB 题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new…

题面 题解 \(Tarjan\)板子题. 统计出大小大于\(1\)的强连通分量数量输出即可. 代码 #include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <algorithm> #include <cmath> #include <cctype> #define gI gi #define itn int #…

题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round Dance. Only cows can perform the Round D…

题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round Dance. Only cows can perform the Round D…

题目描述 约翰的N (2 <= N <= 10,000)只奶牛非常兴奋,因为这是舞会之夜!她们穿上礼服和新鞋子,别 上鲜花,她们要表演圆舞. 只有奶牛才能表演这种圆舞.圆舞需要一些绳索和一个圆形的水池.奶牛们围在池边站好, 顺时针顺序由1到N编号.每只奶牛都面对水池,这样她就能看到其他的每一只奶牛. 为了跳这种圆舞,她们找了 M(2<M< 50000)条绳索.若干只奶牛的蹄上握着绳索的一端, 绳索沿顺时针方绕过水池,另一端则捆在另一些奶牛身上.这样,一些奶牛就可以牵引另一些奶 牛.…

题目链接 赤裸裸的板子,就加一个特判就行.直接上代码 #include<stdio.h> #include<algorithm> #include<iostream> using namespace std; ];//记录入没入栈. ];//特判*1,是强连通分量就直接过了. ]; ];//手写栈. void push(int x)//手写栈ing. { ins[x]=true; stack[++top]=x; return ; } void pop() { ins[s…

题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete with corsages and new shoes. They know that tonight they will each try to perform the Round Dance. Only cows can perform the Round D…

传送门 有向图,找点数大于1的强连通分量个数 ——代码 #include <stack> #include <cstdio> #include <cstring> #include <iostream> ; int n, m, cnt, idx, size, ans; ], next[MAXN << ]; int dfn[MAXN], low[MAXN], belong[MAXN], tot[MAXN]; bool ins[MAXN]; std:…

洛谷 2953 [USACO09OPEN]牛的数字游戏Cow Digit Game 题目描述 Bessie is playing a number game against Farmer John, and she wants you to help her achieve victory. Game i starts with an integer N_i (1 <= N_i <= 1,000,000). Bessie goes first, and then the two players…

P2875 [USACO07FEB]牛的词汇The Cow Lexicon 题目描述 Few know that the cows have their own dictionary with W (1 ≤ W ≤ 600) words, each containing no more 25 of the characters 'a'..'z'. Their cowmunication system, based on mooing, is not very accurate; sometime…

题目简述:一个有向图,求出这个图点数>1的强连通分量的个数. 那么就是tarjan求强联通分量的模板了. 记得要用一个数组标记节点是否在栈中. 1 #include<bits/stdc++.h> 2 using namespace std; 3 const int N=1e5+10; 4 int head[N],nxt[N<<1],to[N<<1],tot; 5 int dfn[N],low[N],st[N],top,idx,cnt,sze[N]; 6 int n…

传送门:https://www.luogu.org/problemnew/show/P2863 思路:tarjan模板题,之前会的tarjan,一直想学缩点到底是什么操作,发现就是把同组的放在一个数组里(并查集可能又会有),或者重新建图:不知道为什么百度不到讲缩点的blog. #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <s…

P1352 没有上司的舞会 题目描述 某大学有N个职员,编号为1~N.他们之间有从属关系,也就是说他们的关系就像一棵以校长为根的树,父结点就是子结点的直接上司.现在有个周年庆宴会,宴会每邀请来一个职员都会增加一定的快乐指数Ri,但是呢,如果某个职员的上司来参加舞会了,那么这个职员就无论如何也不肯来参加舞会了.所以,请你编程计算,邀请哪些职员可以使快乐指数最大,求最大的快乐指数. 输入输出格式 输入格式: 第一行一个整数N.(1<=N<=6000) 接下来N行,第i+1行表示i号职员的快乐指数R…

P3045 [USACO12FEB]牛券Cow Coupons 71通过 248提交 题目提供者洛谷OnlineJudge 标签USACO2012云端 难度提高+/省选- 时空限制1s / 128MB 提交  讨论  题解 最新讨论更多讨论 86分求救 题目描述 Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his budget…

题目描述 Being a secret computer geek, Farmer John labels all of his cows with binary numbers. However, he is a bit superstitious, and only labels cows with binary numbers that have exactly K "1" bits (1 <= K <= 10). The leading bit of each la…

P1569 [USACO11FEB]属牛的抗议Generic Cow Prote- 题目描述 约翰家的N头奶牛聚集在一起,排成一列,正在进行一项抗议活动.第i头奶牛的理智度 为Ai,Ai可能是负数.约翰希望奶牛在抗议时保持理性,为此,他打算将所有的奶牛隔离成 若干个小组,每个小组内的奶牛的理智度总和都要大于等于零.由于奶牛是按直线排列的,所以 一个小组内的奶牛位置必须是连续的. 请帮助约翰计算一下,最多分成几组. 输入输出格式 输入格式: 第1行包含1个数N,代表奶牛的数目. 第2至N+1行每行…

点此看题面 大致题意: 一个由\(R*C\)间矩形宫室组成的宫殿中的\(N\)间宫室里埋藏着宝藏.由一间宫室到达另一间宫室只能通过传送门,且只有埋有宝藏的宫室才有传送门.传送门分为3种,分别可以到达同行的任一宫室(横天门).同列的任一宫室(纵寰门)和以该宫室为中心周围8个的任一宫室(自 由 门).现在你可以从任一宫室开始寻宝,并可以在任一宫室结束寻宝,请求出最多可获得的宝藏数目(每个宝藏只能获得一次). 一个简单的想法 显然,我们可以将每个宫室与它能到达的宫室之间连一条边.由于可能会出现环,我们…

本题是练习前缀和的好题!我们可以枚举前端点,确定一个长度为k的区间,然后利用前缀和统计区间内损坏的灯的数量,最后取最小值即可.AC代码: #include <bits/stdc++.h> using namespace std; inline int read()//快速读入 { ,x=; char c=getchar(); ') { ; c=getchar(); } ') { x=x*+c-'; c=getchar(); } return f*x; } ],s[],sum=; int mai…

题意:n头奶牛,给出若干个欢迎关系a b,表示a欢迎b,欢迎关系是单向的,但是是可以传递的,如:a欢迎b,b欢迎c,那么a欢迎c .另外每个奶牛都是欢迎他自己的.求出被所有的奶牛欢迎的奶牛的数目.#include <iostream>#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <vector> #include &l…