Codeforces Round #335 (Div. 2) 606D Lazy Student(最小生成树+stl)

D. Lazy Student

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Student Vladislav came to his programming exam completely unprepared as usual. He got a question about some strange algorithm on a graph — something that will definitely never be useful in real life. He asked a girl sitting next to him to lend him some cheat papers for this questions and found there the following definition:

The minimum spanning tree T of graph G is such a tree that it contains all the vertices of the original graph G, and the sum of the weights of its edges is the minimum possible among all such trees.

Vladislav drew a graph with n vertices and m edges containing no loops and multiple edges. He found one of its minimum spanning trees and then wrote for each edge its weight and whether it is included in the found tree or not. Unfortunately, the piece of paper where the graph was painted is gone and the teacher is getting very angry and demands to see the original graph. Help Vladislav come up with a graph so that the information about the minimum spanning tree remains correct.

Input

The first line of the input contains two integers n and m () — the number of vertices and the number of edges in the graph.

Each of the next m lines describes an edge of the graph and consists of two integers aj and bj (1 ≤ aj ≤ 109, bj = {0, 1}). The first of these numbers is the weight of the edge and the second number is equal to 1 if this edge was included in the minimum spanning tree found by Vladislav, or 0 if it was not.

It is guaranteed that exactly n - 1 number {bj} are equal to one and exactly m - n + 1 of them are equal to zero.

Output

If Vladislav has made a mistake and such graph doesn't exist, print  - 1.

Otherwise print m lines. On the j-th line print a pair of vertices (uj, vj) (1 ≤ uj, vj ≤ n, uj ≠ vj), that should be connected by the j-th edge. The edges are numbered in the same order as in the input. The graph, determined by these edges, must be connected, contain no loops or multiple edges and its edges with bj = 1 must define the minimum spanning tree. In case there are multiple possible solutions, print any of them.

Sample test(s)

input

4 5
2 1
3 1
4 0
1 1
5 0

output

2 4
1 4
3 4
3 1
3 2

input

3 3
1 0
2 1
3 1

output

-1

题目链接:点击打开链接

n个点, m条边, 给出m条边的权值以及是否在最小生成树中, 如果可以还原最小生成树, 则按照顺序输出各边, 否则输出-1.

考虑这样一种构造方法, 首先对边的权值进行排序, 如果当前边在最小生成树中, 直接将顶点加入. 否则, 为当前边分配顶点, 由于给在最

小生成树的边分配顶点时, 采用了(a, a + 1)这样的分配方式, 所以给不在最小生成树分配顶点时要避免这种方式, 加入判断条件即可.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
int n, m, x, y, z, last;
pair<int, int> ans[MAXN];
vector<pair<pair<int, int>, int > > v;
int main(int argc, char const *argv[])
{
	scanf("%d%d", &n, &m);
	for(int i = 0; i < m; ++i) {
		scanf("%d%d", &x, &y);
		v.push_back(make_pair(make_pair(x, !y), i));
	}
	sort(v.begin(), v.end());
	x = 2, y = 0, last = 1;
	for(int i = 0; i < v.size(); ++i) {
		if(!v[i].first.second) {
			ans[v[i].second] = make_pair(last, last + 1);
			last++;
		}
		else {
			y++;
			if(x == y + 1) {
				x++;
				y = 1;
			}
			if(x > last) {
				printf("-1\n");
				return 0;
			}
			ans[v[i].second] = make_pair(y, x);
		}
	}
	for(int i = 0; i < m; ++i)
		printf("%d %d\n", ans[i].first, ans[i].second);
	return 0;
}