计蒜客 CodeCoder vs TopForces

最新推荐文章于 2020-07-12 08:11:44 发布

什么样的小孩

最新推荐文章于 2020-07-12 08:11:44 发布

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分类专栏： DFS 文章标签： ACM DFS

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本文介绍了一种算法竞赛问题，通过两种不同的编程平台评级系统来确定每位参赛者可能击败的其他参赛者的数量。文章提供了完整的解析思路及实现代码，适用于对算法竞赛感兴趣的学习者。

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CodeCoder vs TopForces

Competitive programming is very popular in Byteland. In fact, every Bytelandian citizen is registered at two programming sites — CodeCoder and TopForces. Each site maintains its own proprietary rating system. Each citizen has a unique integer rating at each site that approximates their skill. Greater rating corresponds to better skill.

People of Byteland are naturally optimistic. Citizen A thinks that he has a chance to beat citizen B in a programming competition if there exists a sequence of Bytelandian citizens A = P0, P1, . . . , Pk = B for some k≥1 such that for each i(0≤i<k),Pi has higher rating than Pi+1 at one or both sites.

Each Bytelandian citizen wants to know how many other citizens they can possibly beat in a programming competition.

Input

The first line of the input contains an integer n — the number of citizens (1 ≤ n ≤ 100000). The following n lines contain information about ratings. The i-th of them contains two integers CCi andT Fi — ratings of the i-th citizen at CodeCoder and TopForces (1 ≤ CCi, T Fi ≤ $10^6$ ). All the ratings at each site are distinct.

Output

For each citizen i output an integer bi — how many other citizens they can possibly beat in a programming competition. Each bi should be printed in a separate line, in the order the citizens are given in the input.

样例输入

样例输出

题意：

每行表示每个人的两个网站的情况，数大的能打败数小的，求每人能打败多少人（只要其中一个网址能赢即可）

思路：

先按第一种排名从小到大排序,
将第i个人与第i+1个人连一条有向边,
表示第i个人能赢第i+1个人;

再按第二种排名从小到大排序,
将第i个人与第i+1个人连一条有向边;

然后按排名从低到高搜索一遍,能搜到的人表示能赢的,
由于传递性,vis不用清空,cnt表示由低到高能赢的人数，
代码最后的cnt-1表示自身也加一，所以减去即为答案

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;

struct data
{
	int u,v;
	int num,cnt;
}p[100005];

vector<int>q[100005];
int vis[100005];
int cnt;

bool cmp_u(data a,data b)
{
	return a.u>b.u;
}

bool cmp_v(data a,data b)
{
	return a.v>b.v;
}

bool cmp_num(data a,data b)
{
	return a.num<b.num;
}

void dfs(int num)
{
	if(!vis[num])cnt++;
	vis[num]=1;
	
	for(int i=0;i<q[num].size();i++)
	{
		int t=q[num][i];
		if(!vis[t]) dfs(t);
	}
}

int  main()
{
	int n;
	cin>>n;
	
	for(int i=0;i<n;i++)
		q[i].clear();
	
	for(int i=0;i<n;i++)
	{
		cin>>p[i].u>>p[i].v;
		p[i].cnt=0;
		p[i].num=i;
	}
		
	sort(p,p+n,cmp_u);
	for(int i=0;i<n-1;i++)
		q[p[i].num].push_back(p[i+1].num);
	
	sort(p,p+n,cmp_v);
	for(int i=0;i<n-1;i++)
		q[p[i].num].push_back(p[i+1].num);
	
	cnt=0;
	memset(vis,0,sizeof(vis));
	for(int i=n-1;i>=0;i--)
	{
		dfs(p[i].num);
		p[i].cnt=cnt-1;
	}
	
	sort(p,p+n,cmp_num);
	for(int i=0;i<n;i++)
		cout<<p[i].cnt<<endl;
		
	return 0;
}