CodeForces 1209 A. Cards with Numbers

A. Cards with Numbers

time limit per test
1 second

memory limit per test
256 megabytes

input
input.txt

output
output.txt

Petya has got 2n cards, each card contains some integer. The numbers on the cards can be the same. Let's index all cards by consecutive integers from 1 to 2n. We'll denote the number that is written on a card with number i, as ai. In order to play one entertaining game with his friends, Petya needs to split the cards into pairs so that each pair had equal numbers on the cards. Help Petya do that.

Input
The first line contains integer n (1 ≤ n ≤ 3·105). The second line contains the sequence of 2n positive integers a1, a2, ..., a2n (1 ≤ ai ≤ 5000) — the numbers that are written on the cards. The numbers on the line are separated by single spaces.

Output
If it is impossible to divide the cards into pairs so that cards in each pair had the same numbers, print on a single line integer -1. But if the required partition exists, then print n pairs of integers, a pair per line — the indices of the cards that form the pairs.

Separate the numbers on the lines by spaces. You can print the pairs and the numbers in the pairs in any order. If there are multiple solutions, print any of them.

Sample test(s)

input

3
20 30 10 30 20 10

output

4 2
1 5
6 3

input

1
1 2

output

-1

解题思路： 虽然有3*10^5个数字，但是大小在[1,5000]之内，对[1,5000]的每一个值出现的位置记录起来
若任意个值出现的次数为奇数则输出-1，否则输出每对就可以了。

View Code

#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;

const int N = 300009;
const int M = 5010;

vector<int> Q[M];
int n, m;
int main()
{
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
while( scanf("%d", &n) != EOF)
{
for(int i = 0; i <= M; i++) Q[i].clear();
for(int i = 1; i <= (n<<1); i++)
{
int x; scanf("%d", &x);
Q[x].push_back(i);
}
bool flag = true;
for(int i = 1; i < M; i++)
{
int cnt = Q[i].size();
if( cnt&1 ){
flag = false;
printf("-1\n");
break;
}
}
if( flag )
{
for(int i = 1; i < M; i++)
{
int cnt = Q[i].size();
for(int j = 0; j < cnt; j += 2)
printf("%d %d\n", Q[i][j], Q[i][j+1] );
}
}
}
return 0;
}