POJ3180:The Cow Prom——题解】的更多相关文章

The Cow Prom Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 2373   Accepted: 1402 Description 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.…
http://poj.org/problem?id=3180 英文题以后都不粘贴题面. 大意:求点数大于1的强连通分量个数 #include<stack> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; typedef long long ll; inline int read(){ ,w=;; ;ch=g…
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() { ;…
http://poj.org/problem?id=3180 The Cow Prom Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 1428   Accepted: 903 Description The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their finest gowns, complete…
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…
1654: [Usaco2006 Jan]The Cow Prom 奶牛舞会 Time Limit: 5 Sec  Memory Limit: 64 MB Description 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 ton…
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 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…
Description 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…
Description 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…
Description The N ( <= N <= ,) 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…
Description 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…
http://www.lydsy.com/JudgeOnline/problem.php?id=1654 请不要被这句话误导..“ 如果两只成功跳圆舞的奶牛有绳索相连,那她们可以同属一个组合.” 这句话没啥用.. #include <cstdio> #include <cstring> #include <cmath> #include <string> #include <iostream> #include <algorithm>…
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…
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…
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…
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]…
[题目链接] http://poj.org/problem?id=3180 [题目大意] N头牛,M条有向绳子,能组成几个歌舞团?要求顺时针逆时针都能带动舞团内所有牛. [题解] 等价于求点数大于1的SCC数量. [代码] #include <cstdio> #include <algorithm> #include <vector> #include <cstring> using namespace std; const int MAX_V=10000;…
2058: [Usaco2010 Nov]Cow Photographs Time Limit: 3 Sec  Memory Limit: 64 MBSubmit: 190  Solved: 104[Submit][Status][Discuss] Description 奶牛的图片 Farmer John希望给他的N(1<=N<=100,000)只奶牛拍照片,这样他就可以向他的朋友炫耀他的奶牛.这N只奶牛被标号为1..N. 在照相的那一天,奶牛们排成了一排.其中第i个位置上是标号为c_i(1…
本蒟蒻又来发题解了, 一道较水的模拟题. 题意不过多解释, 思路如下: 在最开始的时候求出每头牛在t秒的位置(最终位置 然后,如果后一头牛追上了前一头牛,那就无视它, 把它们看成一个整体. else 就++ ans: 上代码: #include<bits/stdc++.h> using namespace std; //要开long long long long n, t, ans = 1, last[100010]; struct node { long long s, p; }a[1000…
P1522 牛的旅行 Cow Tours 题目描述 农民 John的农场里有很多牧区.有的路径连接一些特定的牧区.一片所有连通的牧区称为一个牧场.但是就目前而言,你能看到至少有两个牧区通过任何路径都不连通.这样,Farmer John就有多个牧场了. John想在牧场里添加一条路径(注意,恰好一条).对这条路径有以下限制: 一个牧场的直径就是牧场中最远的两个牧区的距离(本题中所提到的所有距离指的都是最短的距离).考虑如下的有5个牧区的牧场,牧区用"*"表示,路径用直线表示.每一个牧区都…
P1821 [USACO07FEB]银牛派对Silver Cow Party 题目描述 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 co…
想找原题请点击这里:传送门 原题: 题目背景 [Usaco2008 Jan] 题目描述 N ( ≤ N ≤ ) cows, conveniently numbered ..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among…
题目大意:有n个牛在一块, m条单项绳子, 有m个链接关系, 问有多少个团体内部任意两头牛可以相互可达 解题思路:有向图强连通分量模版图 代码如下: #include<stdio.h> #include<vector> #include<map> #include<set> #include<algorithm> using namespace std; typedef long long ll; ; vector<int>G[N],…
裸的强连通 ; type node=record f,t:longint; end; var n,m,dgr,i,u,v,num,ans:longint; bfsdgr,low,head,f:array[..maxe] of longint; b:array[..maxe] of node; p:array[..maxe] of boolean; procedure insert(u,v:longint); begin with b[i] do begin f:=head[u]; t:=v; e…