poj 3180 The Cow Prom(强联通分量)
http://poj.org/problem?id=3180
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 1428 | Accepted: 903 |
Description
Only cows can perform the Round Dance which requires a set of ropes and a circular stock tank. To begin, the cows line up around a circular stock tank and number themselves in clockwise order consecutively from 1..N. Each cow faces the tank so she can see the other dancers.
They then acquire a total of M (2 <= M <= 50,000) ropes all of which are distributed to the cows who hold them in their hooves. Each cow hopes to be given one or more ropes to hold in both her left and right hooves; some cows might be disappointed.
For the Round Dance to succeed for any given cow (say, Bessie), the ropes that she holds must be configured just right. To know if Bessie's dance is successful, one must examine the set of cows holding the other ends of her ropes (if she has any), along with the cows holding the other ends of any ropes they hold, etc. When Bessie dances clockwise around the tank, she must instantly pull all the other cows in her group around clockwise, too. Likewise,
if she dances the other way, she must instantly pull the entire group counterclockwise (anti-clockwise in British English).
Of course, if the ropes are not properly distributed then a set of cows might not form a proper dance group and thus can not succeed at the Round Dance. One way this happens is when only one rope connects two cows. One cow could pull the other in one direction, but could not pull the other direction (since pushing ropes is well-known to be fruitless). Note that the cows must Dance in lock-step: a dangling cow (perhaps with just one rope) that is eventually pulled along disqualifies a group from properly performing the Round Dance since she is not immediately pulled into lockstep with the rest.
Given the ropes and their distribution to cows, how many groups of cows can properly perform the Round Dance? Note that a set of ropes and cows might wrap many times around the stock tank.
Input
Lines 2..M+1: Each line contains two space-separated integers A and B that describe a rope from cow A to cow B in the clockwise direction.
Output
Sample Input
5 4
2 4
3 5
1 2
4 1
Sample Output
1 求有多少个大于2的联通分量
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<vector>
#define N 10010
#define min(a, b)(a < b ? a : b); using namespace std; vector<vector<int> >G; int low[N], dfn[N], Stack[N];
int n, m, num, Time, top;
bool Instack[N]; void Init()
{
G.clear();
G.resize(n + );
memset(low, , sizeof(low));
memset(dfn, , sizeof(dfn));
memset(Stack, , sizeof(Stack));
memset(Instack, false, sizeof(Instack));
Time = top = num = ;
} void Tarjan(int u)
{
int len, v, i, k = ;
low[u] = dfn[u] = ++Time;
Stack[top++] = u;
Instack[u] = true;
len = G[u].size();
for(i = ; i < len ; i++)
{
v = G[u][i];
if(!dfn[v])
{
Tarjan(v);
low[u] = min(low[u], low[v]);
}
else if(Instack[v])
low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u])
{
do
{
k++;
v = Stack[--top];
Instack[v] = false;
}while(v != u);
if(k >= )
num++;
}
} void Solve()
{
int i;
for(i = ; i <= n ; i++)
if(!low[i])
Tarjan(i);
} int main()
{
int a, b;
while(~scanf("%d%d", &n, &m))
{
Init();
while(m--)
{
scanf("%d%d", &a, &b);
G[a].push_back(b);
}
Solve();
printf("%d\n", num);
}
return ;
}
poj 3180 The Cow Prom(强联通分量)的更多相关文章
- POJ 3180 The Cow Prom(强联通)
题目大意: 约翰的N(2≤N≤10000)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 ...
- POJ 2186 Popular cows(Kosaraju+强联通分量模板)
题目链接:http://poj.org/problem?id=2186 题目大意:给定N头牛和M个有序对(A,B),(A,B)表示A牛认为B牛是红人,该关系具有传递性,如果牛A认为牛B是红人,牛B认为 ...
- POJ 1904 King's Quest 强联通分量+输入输出外挂
题意:国王有n个儿子,现在这n个儿子要在n个女孩里选择自己喜欢的,有的儿子可能喜欢多个,最后国王的向导给出他一个匹配.匹配有n个数,代表某个儿子和哪个女孩可以结婚.已知这些条件,要你找出每个儿子可以和 ...
- POJ 3180 The cow Prom Tarjan基础题
题目用google翻译实在看不懂 其实题目意思如下 给一个有向图,求点个数大于1的强联通分量个数 #include<cstdio> #include<algorithm> #i ...
- POJ 3592 Instantaneous Transference(强联通分量 Tarjan)
http://poj.org/problem?id=3592 题意 :给你一个n*m的矩阵,每个位置上都有一个字符,如果是数字代表这个地方有该数量的金矿,如果是*代表这个地方有传送带并且没有金矿,可以 ...
- poj 3180 The Cow Prom(tarjan+缩点 easy)
Description The N ( <= N <= ,) cows are so excited: it's prom night! They are dressed in their ...
- POJ 3180 The Cow Prom(SCC)
[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 ...
- [poj] 3180 the cow prom
原题 这是一道强连通分量板子题. 我们只用输出点数大于1的强连通分量的个数! #include<cstdio> #include<algorithm> #include< ...
- USACO06JAN The Cow Prom /// tarjan求强联通分量 oj24219
题目大意: n个点 m条边的图 求大小大于1的强联通分量的个数 https://www.cnblogs.com/stxy-ferryman/p/7779347.html tarjan求完强联通分量并染 ...
随机推荐
- C# 按拼音/笔划 排序的简单示例(转)
class Program { static void Main(string[] args) { string[] arr = { "趙(ZHAO)", "錢(QIAN ...
- Git show-branch显示提交信息
git中查看日志,我们用的比较多的就是 git log 以及带一些参数,如: 以一行显示提交日志: $ git log --pretty=oneline 显示最后的几次提交日志: $ git log ...
- ie6调试工具Debugbar
http://www.my-debugbar.com/wiki/Doc/DebugbarInstall
- TCP建立连接和释放的过程,及TCP状态变迁图
一.TCP报文格式 下面是TCP报文格式图: 重要字段介绍: (1)序号:Seq序号,占32位,用来标识从TCP源端向目的端发送的字节流,发起方发送数据时对此进行标记. (2)确认序号:Ack序号,占 ...
- Topcoder SRM 630 (500 floyed 暴力 _builtin_popcount())
题意:给n个点,保证图联通,给点相连的距离,求一个最多的点,这些点之间的距离都是相同的. 分析: 下面的代码是我们房间第一的大神的,写的很简洁,我的思路和他的一样,但是我不知道错哪了. 思路是暴力枚举 ...
- bzoj1221: [HNOI2001] 软件开发
挖坑.我的那种建图方式应该也是合理的.然后连样例都过不了.果断意识到应该为神奇建图法... #include<cstdio> #include<cstring> #includ ...
- Test语言编译器V0.8
感觉这个挺好耍的,书上的代码有错误,而且功能有限. 一.词法分析 特点: (1)可对中文进行识别:(2)暂不支持负数,可以在读入‘-'时进行简单标记后就能对简单负数进行识别了. #include &l ...
- 转:MVC3系列:~Html.BeginForm与Ajax.BeginForm
Html.BeginForm与Ajax.BeginForm都是MVC架构中的表单元素,它们从字面上可以看到区别,即Html.BeginForm是普通的表单提交,而Ajax.BeginForm是支持异步 ...
- Material Design 设计--阴影的重要性
<LinearLayout android:layout_width="match_parent" android:layout_height="wrap_cont ...
- UVA 11294 Wedding(2-sat)
2-sat.不错的一道题,学到了不少. 需要注意这么几点: 1.题目中描述的是有n对夫妇,其中(n-1)对是来为余下的一对办婚礼的,所以新娘只有一位. 2.2-sat问题是根据必然性建边,比如说A与B ...