Codeforces Round #599 (Div. 2)D 边很多的只有0和1的MST
题:https://codeforces.com/contest/1243/problem/D
分析:找全部可以用边权为0的点连起来的全部块
然后这些块之间相连肯定得通过边权为1的边进行连接
所以答案就是这些块的总数-1;
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
const int M=1e5+;
set<int>s,g[M];
int vis[M];
void bfs(int x){
queue<int>que;
que.push(x);
s.erase(x);
while(!que.empty()){
int u=que.front();
que.pop();
if(vis[u])
continue;
vis[u]=;
set<int>::iterator it;
for(it=s.begin();it!=s.end();){
int v=*it;
it++;
if(g[u].find(v)==g[u].end()){///如果这一次找不到的话,就说明这个块中的点有流向它的权值为1的边
que.push(v);
s.erase(v); }
}
}
}
int main(){
int n,m;
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
s.insert(i);
for(int x,y,i=;i<=m;i++){
scanf("%d%d",&x,&y);
g[x].insert(y);
g[y].insert(x);
}
int ans=;
for(int i=;i<=n;i++){
if(!vis[i]){
ans++;
bfs(i);
}
}
printf("%d\n",ans-);
}
Codeforces Round #599 (Div. 2)D 边很多的只有0和1的MST的更多相关文章
- Codeforces Round #599 (Div. 2) D. 0-1 MST(bfs+set)
Codeforces Round #599 (Div. 2) D. 0-1 MST Description Ujan has a lot of useless stuff in his drawers ...
- Codeforces Round #367 (Div. 2) D. Vasiliy's Multiset (0/1-Trie树)
Vasiliy's Multiset 题目链接: http://codeforces.com/contest/706/problem/D Description Author has gone out ...
- Codeforces Round #599 (Div. 2)
久违的写篇博客吧 A. Maximum Square 题目链接:https://codeforces.com/contest/1243/problem/A 题意: 给定n个栅栏,对这n个栅栏进行任意排 ...
- Codeforces Round #599 (Div. 2) E. Sum Balance
这题写起来真的有点麻烦,按照官方题解的写法 先建图,然后求强连通分量,然后判断掉不符合条件的换 最后做dp转移即可 虽然看起来复杂度很高,但是n只有15,所以问题不大 #include <ios ...
- Codeforces Round #599 (Div. 1) C. Sum Balance 图论 dp
C. Sum Balance Ujan has a lot of numbers in his boxes. He likes order and balance, so he decided to ...
- Codeforces Round #599 (Div. 1) B. 0-1 MST 图论
D. 0-1 MST Ujan has a lot of useless stuff in his drawers, a considerable part of which are his math ...
- Codeforces Round #599 (Div. 1) A. Tile Painting 数论
C. Tile Painting Ujan has been lazy lately, but now has decided to bring his yard to good shape. Fir ...
- Codeforces Round #599 (Div. 2) B2. Character Swap (Hard Version) 构造
B2. Character Swap (Hard Version) This problem is different from the easy version. In this version U ...
- Codeforces Round #599 (Div. 2) B1. Character Swap (Easy Version) 水题
B1. Character Swap (Easy Version) This problem is different from the hard version. In this version U ...
随机推荐
- [极客大挑战 2019]FinalSQL
0x00 知识点 盲注 0x01 解题 根据题目提示盲注,随便点几下找到注入点 发现我们输入^符号成功跳转页面,证明存在注入 1^(ord(substr((select(group_concat(sc ...
- Zookeeper--命令介绍
参考 https://zookeeper.apache.org/doc/r3.4.13/zookeeperStarted.html#sc_ConnectingToZooKeeper 连接到zookee ...
- openstack trove主要贡献公司-Tesora被Stratoscale收购
新闻链接:http://www.stratoscale.com/press/press-releases/stratoscale-acquires-database-as-a-service-prov ...
- 【收藏】免费开源好看的bootstrap后台模板
1.ace admin github:https://github.com/bopoda/acedemo:http://ace.jeka.by/ 2.CoreUI jQuery.Angular.Rea ...
- 【十日冲刺计划】第一日 星遇Sprint1计划会议成果
——小组成员 赵剑峰 张傲 周龙海 Sprint 计划会议1:定出 Sprint 目标和既定产品Backlog. 会议进程(1小时) • 首先对sprint目标进行总体介绍,概括星遇的backlog, ...
- 对比Node.js和Python 帮你确定理想编程解决方案!
世上没有最好的编程语言.有些编程语言比其他编程语言用于更具体的事情.比如,你可能需要移动应用程序,网络应用程序或更专业化的系统,则可能会有特定的语言.但是我们暂时假设你需要的是一个相对来说比较简单的网 ...
- E. Double Elimination (DP)
题目:传送门 题意:有 2^n 个人进行比赛,对他们编号 1~2^n,起初1和2打,3和4打,5和6打,7和8打...,然后1和2打完胜利的 和 3和4打完胜利的再打一场,1和2打完失败的和3和4打完 ...
- 吴裕雄--天生自然 JAVASCRIPT开发学习:(String) 对象
<!DOCTYPE html> <html> <head> <meta charset="utf-8"> <title> ...
- 0CTF-2016-piapiapia-PHP反序列化长度变化尾部字符串逃逸
0X00 扫描一下网站目录,得到网站源码,这里说下工具使用的是dirmap,亲测御剑不好用... 0x01 审计源码: index.php <?php require_once('class.p ...
- Linux 杀掉所有Java进程
ps -ef | grep java | grep -v grep | awk '{print $2}' | xargs kill -9 管道符"|"用来隔开两个命令,管道符左 ...