D. Lazy Student
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/606/problem/D
Description
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 Input
4 5
2 1
3 1
4 0
1 1
5 0
Sample Output
2 4
1 4
3 4
3 1
3 2
HINT
题意
有一个n点m边的图，没有自环，没有重边
然后给你这m条边的大小，以及哪些边属于一颗最小生成树里面的
现在让你构造一个图，使得，满足对应的边，确实属于一颗最小生成树里面。
如果不能构造，输出-1.
题解：
我们首先把所有边都读入，然后按照边权从小到大排序，边权一样，树边优先。
然后依次插入。
如果插入的是树边，就直接插入就好了。
如果插入的是图边，那么图边一定比他所在环的所有树边的边权都大。
把树边插成一个菊花图/链图都可以
然后图边就直接插入就好了，插入之后，控制l++，因为r不变的话，图边的边权要求也是不会变的。
如果l == r的话，看是否r+1这个点被树边插过，如果没有就return -1
否则就r++,l = 1就行了
一直循环就好了
代码：
#include<iostream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<algorithm>
using namespace std;
#define maxn 100005 struct node
{
int x,y,z;
};
bool cmp(node A,node B)
{
if(A.x==B.x)
return A.y>B.y;
return A.x<B.x;
}
int l = ,r = ;
vector<node> p;
int ans1[maxn],ans2[maxn];
int vis[maxn];
int main()
{
int n,m;
cin>>n>>m;
for(int i=;i<=m;i++)
{
node k;
scanf("%d%d",&k.x,&k.y);
k.z = i;
p.push_back(k);
}
sort(p.begin(),p.end(),cmp);
int rr = ;
for(int i=;i<p.size();i++)
{
if(p[i].y==)
{
ans1[p[i].z]=,ans2[p[i].z]=rr;
vis[rr]=;rr++;
}
else if(p[i].y==)
{
if(l==r)
{
r++;l=;
if(vis[r]==)return puts("-1");
}
ans1[p[i].z]=l++,ans2[p[i].z]=r;
}
}
for(int i=;i<=m;i++)
printf("%d %d\n",ans1[i],ans2[i]);
}