AtCoder Beginner Contest 113 C

C - ID

Time limit : 2sec / Memory limit : 1024MB

Score: 300 points

Problem Statement

In Republic of Atcoder, there are N prefectures, and a total of M cities that belong to those prefectures.

City i is established in year Yi and belongs to Prefecture Pi.

You can assume that there are no multiple cities that are established in the same year.

It is decided to allocate a 12-digit ID number to each city.

If City i is the x-th established city among the cities that belong to Prefecture i, the first six digits of the ID number of City i is Pi, and the last six digits of the ID number is x.

Here, if Pi or x (or both) has less than six digits, zeros are added to the left until it has six digits.

Find the ID numbers for all the cities.

Note that there can be a prefecture with no cities.

Constraints

1≤N≤105
1≤M≤105
1≤Pi≤N
1≤Yi≤109
Yi are all different.
All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

P1 Y1

:

PM YM

Output

Print the ID numbers for all the cities, in ascending order of indices (City 1, City 2, …).

Sample Input 1

Copy

Sample Output 1

Copy

000001000002

000002000001

000001000001

As City 1 is the second established city among the cities that belong to Prefecture 1, its ID number is 000001000002.
As City 2 is the first established city among the cities that belong to Prefecture 2, its ID number is 000002000001.
As City 3 is the first established city among the cities that belong to Prefecture 1, its ID number is 000001000001.

Sample Input 2

Copy

Sample Output 2

Copy

000002000001

000002000002

000002000003

高级点的结构体排序，主要难点是判断每个区间的数字排序

#include <iostream>

#include <algorithm>

#include <cstring>

#include <cstdio>

#include <vector>

#include <queue>

#include <stack>

#include <cstdlib>

#include <iomanip>

#include <cmath>

#include <cassert>

#include <ctime>

#include <cstdlib>

#include <map>

#include <set>

using namespace std;

#pragma comment(linker, "/stck:1024000000,1024000000")

#define lowbit(x) (x&(-x))

#define max(x,y) (x>=y?x:y)

#define min(x,y) (x<=y?x:y)

//#define MAX 100000000000000000

#define MOD 1000000007

#define pi acos(-1.0)

#define ei exp(1)

#define PI 3.1415926535897932384626433832

#define ios() ios::sync_with_stdio(true)

#define INF (1<<31)-1;

#define mem(a) (memset(a,0,sizeof(a)))

struct P{

    int x,y;

}H[];

vector<int>G[];

int POS(int k,int value){

    int mid;

    int left = ;

    int len = G[k].size();

    int right = len - ;

    //cout<<" "<<k<<"A"<<value<<endl;

    while(left <= right){

        // 确保中点靠近区间的起点

        mid = left + (right-left)/;

        //cout<<mid<<endl;

        //cout<<"mid= "<<mid<<" "<<left<<" "<<right<<endl;

        // 如果找到则返回

        if(G[k][mid] == value) return mid;

        // 将中点赋给终点

        else if(G[k][mid] > value) right = mid;

        // 将中点加一赋给起点

        else left = mid + ;

    }

    return -;

}

set<int>::iterator it;

int main()

{

    set<int>s;

    int N,M;

    int x,y;

    cin>>N>>M;

    for(int i=;i<=M;i++){

        cin>>x>>y;

        H[i].x = x;

        H[i].y = y;

        G[x].push_back(y);

        s.insert(x);

    }

    for(int i=;i<=N;i++){

        sort(G[i].begin(),G[i].end());

    }

    for(int i=;i<=M;i++){

        cout<<setw()<<setfill('')<<H[i].x;

        cout<<setw()<<setfill('')<<POS(H[i].x,H[i].y)+<<endl;

    }

    return ;

}