[luoguP2863] [USACO06JAN]牛的舞会The Cow Prom(Tarjan)
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> const int MAXN = ;
int n, m, cnt, idx, size, ans;
int head[MAXN], to[MAXN << ], next[MAXN << ];
int dfn[MAXN], low[MAXN], belong[MAXN], tot[MAXN];
bool ins[MAXN];
std::stack <int> s; inline int read()
{
int x = , f = ;
char ch = getchar();
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -;
for(; isdigit(ch); ch = getchar()) x = (x << ) + (x << ) + ch - '';
return x * f;
} inline int min(int x, int y)
{
return x < y ? x : y;
} inline void add(int x, int y)
{
to[cnt] = y;
next[cnt] = head[x];
head[x] = cnt++;
} inline void dfs(int u)
{
int i, v;
s.push(u);
ins[u] = ;
dfn[u] = low[u] = ++idx;
for(i = head[u]; i ^ -; i = next[i])
{
v = to[i];
if(!dfn[v])
{
dfs(v);
low[u] = min(low[u], low[v]);
}
else if(ins[v]) low[u] = min(low[u], dfn[v]);
}
if(low[u] == dfn[u])
{
size++;
do
{
v = s.top();
s.pop();
ins[v] = ;
belong[v] = size;
}
while(u ^ v);
}
} int main()
{
int i, x, y;
n = read();
m = read();
memset(head, -, sizeof(head));
for(i = ; i <= m; i++)
{
x = read();
y = read();
add(x, y);
}
for(i = ; i <= n; i++)
if(!dfn[i])
dfs(i);
for(i = ; i <= n; i++) tot[belong[i]]++;
for(i = ; i <= size; i++)
if(tot[i] > )
ans++;
printf("%d\n", ans);
return ;
}
[luoguP2863] [USACO06JAN]牛的舞会The Cow Prom(Tarjan)的更多相关文章
- luoguP2863 [USACO06JAN]牛的舞会The Cow Prom
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 1 ...
- [USACO06JAN]牛的舞会The Cow Prom Tarjan
题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their ...
- luogu P2863 [USACO06JAN]牛的舞会The Cow Prom |Tarjan
题目描述 The N (2 <= N <= 10,000) cows are so excited: it's prom night! They are dressed in their ...
- bzoj1654 / P2863 [USACO06JAN]牛的舞会The Cow Prom
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&l ...
- P2863 [USACO06JAN]牛的舞会The Cow Prom
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 ...
- [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 ...
- 洛谷——P2863 [USACO06JAN]牛的舞会The Cow Prom
https://www.luogu.org/problem/show?pid=2863#sub 题目描述 The N (2 <= N <= 10,000) cows are so exci ...
- [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 ...
- 洛谷 P2863 [USACO06JAN]牛的舞会The Cow Prom
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&g ...
随机推荐
- vue用户登录状态判断
之前项目中用来判断是否登录我写了多种方案,但是最终只有一个方案是比较好的,这篇博客就是分享该方案; 先说基本要求: 项目中的登录状态是依据服务器里的状态来作为判断依据; 每一个需要登录后才能操作的接口 ...
- mysql left join 出现的结果会重复
left join 基本用法 MySQL left join 语句格式 A LEFT JOIN B ON 条件表达式 left join 是以A表为基础,A表即左表,B表即右表. 左表(A)的记录会全 ...
- [USACO09NOV]灯Lights
题目描述 Bessie and the cows were playing games in the barn, but the power was reset and the lights were ...
- 简单几何(水)BestCoder Round #50 (div.2) 1002 Run
题目传送门 /* 好吧,我不是地球人,这题只要判断正方形就行了,正三角形和正五边形和正六边形都不可能(点是整数). 但是,如果不是整数,那么该怎么做呢?是否就此开启计算几何专题了呢 */ /***** ...
- 小白的python之路 序
计算机专科毕业,.net开发已有8年有余,中途断断续续,似懂非懂,积累了一些经验知识,但是不求甚解,属于那种一瓶不满半瓶子晃荡,这么一个状态. 主要从事web开发,涉及一些前端jq等,还有接口开发,搜 ...
- Java 8 (2) 使用Lambda表达式
什么是Lambda? 可以把Lambda表达式理解为 简洁的表示可传递的匿名函数的一种方式:它没有名称,但它有参数列表.函数主体.返回类型,可能还有一个可以抛出的异常列表. 使用Lambda可以让你更 ...
- jsTree插件简介(三)
UI-plugin JSTree的UI插件:用来处理选择.不选和鼠标悬浮树选项的插件. 一.属性包括: 1.select_limit:指定一次可以选中几个节点,默认为-1,表示无限制选中. 2.sel ...
- iOS App Crash原理分析
预备知识:OS X系统分析 1.内核XNU是Darwin的核心,也是整个OS X的核心.XNU本身由以下几个组件构成: Mach微核心 BSD层 libKern I/O Kit 此外,内核是模块化的, ...
- sort 及lambda表达式定制排序
void CDrawCircle::OnBnClickedBtnSelectinfo() { // TODO: 在此添加控件通知处理程序代码 UpdateData(TRUE); BeginEditor ...
- 微服务网关从零搭建——(七)更改存储方式为oracle
资源准备: 下载开源项目 新建oracle表: -- ---------------------------- -- Table structure for OcelotGlobalConfigura ...