Codeforces Round #335 (Div. 2)D. Lazy Student

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#Codeforces Round #33

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探讨了如何根据已知的边权值和最小生成树信息重建原始图的问题，确保重建后的图能形成正确的最小生成树。

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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 ( $\text{[math]}$ ) — 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.

Examples

input
5
1
1
0
1
0

output
4
4
4
1
2

input
3
0
1
1

output
-1

题意：n个顶点m条边，对于每条边给出它的权值和这条边是否是最小生成树里面的边。数据保证所给的最小生成树里的边数为n-1条问能否由这些边组最小生成树。如果不能输出-1否则输出每条边对应的顶点。

思路：将每条边的权值从小到大排序并保证最小生成树里的边在最前面然后最前面的n-1 条边的顶点依次对应为1~n剩下的边依次判断加入即可。

/* ***********************************************
Author       : ryc
Created Time : 2016-08-22 Monday
File Name    : E:\acm\codeforces\335D.cpp
Language     : c++
Copyright 2016 ryc All Rights Reserved
************************************************ */
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<list>
#include<vector>
#include<stack>
#include<map>
using namespace std;
typedef long long LL;
typedef pair<int,int>pii;
const int maxn=100010;
struct Edge{
    int id,w,u,v,b;
}E[maxn];
bool cmp1(Edge a,Edge b){
    if(a.b==b.b)
        return a.w<b.w;
    return a.b>b.b;
}
bool cmp2(Edge a,Edge b){
    return a.id<b.id;
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i){
        scanf("%d%d",&E[i].w,&E[i].b);
        E[i].id=i;
    }
    sort(E+1,E+m+1,cmp1);
    for(int i=1;i<n;++i){
        E[i].u=i;E[i].v=i+1;//前n条边顶点分别为i和i+1
    }
    int f=3,t=0;
    for(int i=n;i<=m;++i){
        t++;if(t+2>f){f++;t=1;}
        if(E[i].w<E[f-1].w){printf("-1\n");return 0;}//保证当前又加入的边一定要大于由顶点1到这个顶点的最小生成树上的边即可
        E[i].u=f;E[i].v=t;
    }
    sort(E+1,E+m+1,cmp2);
    for(int i=1;i<=m;++i){
        printf("%d %d\n",E[i].u,E[i].v);
    }
    return 0;
}