计蒜客 CodeCoder vs TopForces

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. CitizenA thinks that he has a chance to beat citizenB in a programming competition if there exists a sequence of Bytelandian citizensA = P0, P1, . . . , Pk= B for some k≥1 such that for each i(0≤i<k),Pihas higher rating than Pi+1at 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 integern — the number of citizens (1≤ n ≤ 100000). The followingn lines contain information about ratings. Thei-th of them contains two integersCCi andT Fi— ratings of the i-th citizen at CodeCoder and TopForces (1≤ CCi, T Fi≤ 10610^610​6​​). All the ratings at each site are distinct.

Output

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

样例输入

42 33 21 14 5

样例输出

2203

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;}