Codeforces Round #246 (Div. 2) —B. Football Kit

B. Football Kit

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider a football tournament where n teams participate. Each team has two football kits: for home games, and for away games. The kit for home games of the i-th team has color xi and the kit for away games of this team has color yi (xi ≠ yi).

In the tournament, each team plays exactly one home game and exactly one away game with each other team (n(n - 1) games in total). The team, that plays the home game, traditionally plays in its home kit. The team that plays an away game plays in its away kit. However, if two teams has the kits of the same color, they cannot be distinguished. In this case the away team plays in its home kit.

Calculate how many games in the described tournament each team plays in its home kit and how many games it plays in its away kit.

Input

The first line contains a single integer n (2 ≤ n ≤ 105) — the number of teams. Next n lines contain the description of the teams. The i-th line contains two space-separated numbers xi, yi (1 ≤ xi, yi ≤ 105; xi ≠ yi) — the color numbers for the home and away kits of the i-th team.

Output

For each team, print on a single line two space-separated integers — the number of games this team is going to play in home and away kits, correspondingly. Print the answers for the teams in the order they appeared in the input.

Sample test(s)

input

21 22 1

output

2 02 0

input

31 22 11 3

output

3 14 02 2

题意：打印当前队 穿主场衣服作战的次数 和  穿客场衣服作战的次数

思路：简单的哈希思想+规律  即可AC，只是 题意不是很好理解

规律是 n个队  每个队会进行 n-1次 比赛  主场服使用次数 + 客场服使用次数  = 2*（n-1）

主场服使用次数  =  n-1 + 当前队客场服与其他队 主场服相同的次数

#include <stdio.h>#include <stdlib.h>#include <string.h>int a[100010];int b[100010],c[100010];int main(){    int n,i;    scanf("%d",&n);    memset(a,0,sizeof(a));    for(i = 0; i<n; i++)    {        scanf("%d%d",&b[i],&c[i]);        a[b[i]]++;    }    for(i = 0; i<n; i++)    {        printf("%d %d\n",n-1+a[c[i]],2*(n-1)-(n-1+a[c[i]]));    }    return 0;}