CF - 605B.Lazy Student，构造最小生成树

B. 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 52 13 14 01 15 0

output

2 41 43 43 13 2

input

3 31 02 13 1

output

-1

题目大意：

    一张图，n个顶点m条边，只给出它们的权重和是否是最小生成树的边，恢复原来的顶点的连接关系。

解题思路：

    构造题，把最小生成树当成长度为n的链，且是从小到大排序的，于是后面的不是最小生成树的边的两点就只能在在当前这个这个顶点的前面。注意不要有重边。

AC代码：

#include <bits/stdc++.h>#define INF 0x7fffffff#define maxn 1001000#define eps 1e-6#define pi acos(-1.0)#define e 2.718281828459#define mod (int)1e9 + 7using namespace std;typedef long long ll;struct node{  int x, y, v, id, flag;} p[maxn];int cnt[maxn]; // 记录树的情况bool cmp1(node a, node b){ //把最小生成树当成长度为n的链，且是从小到大排序的，于是后面的不是最小生成树的边的两点就只能在在当前这个这个顶点的前面。  if (a.v == b.v)    return a.flag > b.flag;  return a.v < b.v;}bool cmp2(node a, node b){  return a.id < b.id;}int main(){#ifndef ONLINE_JUDGE  freopen("in.txt", "r", stdin);  freopen("out.txt", "w", stdout);  long _begin_time = clock();#endif  int n, m;  while (scanf("%d%d", &n, &m) != EOF)  {    for (int i = 0; i < m; i++)    {      scanf("%d%d", &p[i].v, &p[i].flag);      p[i].id = i;    }    sort(p, p + m, cmp1);    int flag = 1;    if (p[0].flag != 0)    {      int now = 2; //当前最小生成树的顶点      int cur = 3; //不是最小生成树的顶点，它只能在最小生成树前面      cnt[cur] = 1;      for (int i = 0; i < m; i++)      {        if (p[i].flag)        {          p[i].x = now;          p[i].y = now - 1;          now++;        }        else        {          if (cur <= now - 1) //不是最小生成树，那么它的边所在的两个点只能在当前最小生成树两点之前出现          {            p[i].x = cur;            p[i].y = cnt[cur];            if (cnt[cur] >= cur - 2) // cur 和cur + 1的边是最小生成树              cnt[++cur] = 1;            else              cnt[cur]++;          }          else          {            flag = 0;            break;          }        }      }      if (flag)      {        sort(p, p + m, cmp2);        for (int i = 0; i < m; i++)          printf("%d %d\n", p[i].x, p[i].y); //cf的任意输出卡了很久      }      else        puts("-1");    }    else      puts("-1");  }#ifndef ONLINE_JUDGE  long _end_time = clock();  printf("time = %ld ms.", _end_time - _begin_time);#endif  return 0;}