K. Topological Sort

题面

Consider a directed graph G of N nodes and all edges (u→v) such that u < v. It is clear that this graph doesn’t contain any cycles.

Your task is to find the lexicographically largest topological sort of the graph after removing a given list of edges.

A topological sort of a directed graph is a sequence that contains all nodes from 1 to N in some order such that each node appears in the sequence before all nodes reachable from it.

Input

The first line of input contains a single integer T, the number of test cases.

The first line of each test case contains two integers N and M (1 ≤ N ≤ 105) , the number of nodes and the number of edges to be removed, respectively.

Each of the next M lines contains two integers a and b (1 ≤ a < b ≤ N), and represents an edge that should be removed from the graph.

No edge will appear in the list more than once.

Output

For each test case, print N space-separated integers that represent the lexicographically largest topological sort of the graph after removing the given list of edges.

题意

给出N个点,每个对于每个对(u,v)若u<v,那么连一条边,给出m条边,先删除他们,问拓扑序字典序最大是多少。

考场上想了半天,大概最后想用最长路做,但是搞不定。。然后看题解嘛、。。 用线段树维护,操作有:区间修改 单点修改 查询最靠右0的位置。

代码

#include <bits/stdc++.h>
using namespace std; int T;
int n;
int m;
int x,y;
vector<int> g[100010];
int a[4000010];
int lazy[4000010]; void build(int l,int r,int o)
{
if (l==r)
{
a[o]=l-1;
return;
}
int mid=(l+r)>>1;
build(l,mid,o<<1);
build(mid+1,r,o<<1|1);
a[o]=min(a[o<<1],a[o<<1|1]);
return;
} void push_down(int o)
{
lazy[o<<1]+=lazy[o];
lazy[o<<1|1]+=lazy[o];
a[o<<1]+=lazy[o];
a[o<<1|1]+=lazy[o];
lazy[o]=0;
} void add(int L,int R,int l,int r,int o,int dlt)
{
if (L<=l && R>=r)
{
lazy[o]+=dlt;
a[o]+=dlt;
return;
}
push_down(o);
int mid=(l+r)>>1;
if (L<=mid) add(L,R,l,mid,o<<1,dlt);
if (R>mid) add(L,R,mid+1,r,o<<1|1,dlt);
a[o]=min(a[o<<1],a[o<<1|1]);
} int query(int l,int r,int o)
{
if (l==r) return l;
push_down(o);
int mid=(l+r)>>1;
if (a[o<<1|1]==0) return query(mid+1,r,o<<1|1);
else return query(l,mid,o<<1);
} int main()
{
ios::sync_with_stdio(false);
cin>>T;
while (T--)
{
cin>>n>>m;
memset(a,0,sizeof a);
memset(lazy,0,sizeof lazy);
build(1,n,1);
for (int i=1;i<=m;i++)
{
cin>>x>>y;
g[x].push_back(y);
add(y,y,1,n,1,-1);
}
for (int i=1;i<n;i++)
{
int find=query(1,n,1);
cout<<find<<" ";
if (find+1<=n) add(find+1,n,1,n,1,-1);
for (int o=0;o<g[find].size();o++) add(g[find][o],g[find][o],1,n,1,1);
add(find,find,1,n,1,2*n);
}
int find=query(1,n,1);
cout<<find<<endl;
for (int i=1;i<=n;i++) g[i].clear();
}
}

题目链接

http://codeforces.com/gym/101102/problem/K

Gym-101102-K-Topological Sort的更多相关文章

  1. 【拓扑排序】【线段树】Gym - 101102K - Topological Sort

    Consider a directed graph G of N nodes and all edges (u→v) such that u < v. It is clear that this ...

  2. topological sort~~~~初学

    今天讲了topological sort 问题: 判环:记录入队的点数,若<n则有环,可证: 算法:o(n):queue or  stack,而不是o(n^2)枚举 #. 关系运算图(vijos ...

  3. Some facts about topological sort

    Definition: a topological sort of a DAG G is a sort such that for all edge (i,j) in G, i precedes j. ...

  4. topological sort

    A topological sortof a dag G  is a linear ordering of all its vertices such that if G contains anedg ...

  5. 拓扑排序(Topological Sort)

    Graph 拓扑排序(Topological Sort) 假设一个应用场景:你用 C 编写了一个爬虫工具,其中有很多自定义的库:queue.c.queue.h.stack.c.stack.h.heap ...

  6. 6-16 Topological Sort(25 分)

    Write a program to find the topological order in a digraph. Format of functions: bool TopSort( LGrap ...

  7. [Algorithms] Topological Sort

    Topological sort is an important application of DFS in directed acyclic graphs (DAG). For each edge ...

  8. gym/102021/K GCPC18 背包dp算不同数和的可能

    gym/102021/K 题意: 给定n(n<=60)个直线 ,长度<=1000; 可以转化为取 计算 ans = (sum  + 10 - g) / ( n + 1)  在小于5的条件下 ...

  9. [MIT6.006] 14. Depth-First Search (DFS), Topological Sort 深度优先搜索,拓扑排序

    一.深度优先搜索 它的定义是:递归探索图,必要时要回溯,同时避免重复. 关于深度优先搜索的伪代码如下: 左边DFS-Visit(V, Adj.s)是只实现visit所有连接某个特定点(例如s)的其他点 ...

  10. Gym - 100283K K. Cubes Shuffling —— 贪心

    题目链接:http://codeforces.com/gym/100283/problem/K 题解: 要使其相邻两项的差值之和最小,那么越靠中间,其数值越小. 那么剩下的问题就是如何放数字了.一开始 ...

随机推荐

  1. 我的第一个博客——Fragment遇到的问题

    最近项目中使用fragment时遇到了一些问题: 1.fragment的刷新问题. 解决:我的情况是有多个fragment时,只需要刷新其中几个界面.之前我在网上看到的一些方法.如下: 首先在Adap ...

  2. maven工程的拆分与聚合

    1.创建父工程: 首先创建maven Project: 点击next,进入下一个页面,勾选上create a simple project: 点击next,进入下一页面,在这里与创建普通maven工程 ...

  3. Spring Boot学习笔记2——基本使用之最佳实践[z]

    前言 在上一篇文章Spring Boot 学习笔记1——初体验之3分钟启动你的Web应用已经对Spring Boot的基本体系与基本使用进行了学习,本文主要目的是更加进一步的来说明对于Spring B ...

  4. Vue 局部组件和全局组件的使用

    <template> <div id="app"> <!--<img alt="Vue logo" src="./ ...

  5. python 数据类型 之 字典

    python 3.6.5字典的特性和定义定义:{'key_1':vlaue_1,'key_2':value_2}1.键与值用冒号 : 分开2.项与项 用 , 分开 特性1.可以存放多个值,可以不唯一, ...

  6. pyspider入门

    1.http://www.pyspider.cn/jiaocheng/pyspider-webui-12.html 2.https://blog.csdn.net/weixin_37947156/ar ...

  7. js函数定义和调用

    由于JavaScript的函数也是一个对象,上述定义的abs()函数实际上是一个函数对象,而函数名abs可以视为指向该函数的变量. var abs = function (x) { if (x > ...

  8. HTML&&css练习

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  9. JTable 查询

    public JTable query(String table) throws SQLException { DefaultTableModel tablemodel = new DefaultTa ...

  10. 探索未知种族之osg类生物---呼吸分解之advance

    回顾 我们用了两节的内容才堪堪讲解完ViewerBase::frame()函数中调用的realize()---Viewer:: realize()函数.我们简单的总结就是Viewer:: realiz ...