CodeForces - 404C Restore Graph

Valera had an undirected connected graph without self-loops and multiple edges consisting of n vertices. The graph had an interesting property: there were at most k edges adjacent to each of its vertices. For convenience, we will assume that the graph vertices were indexed by integers from 1 to n.

One day Valera counted the shortest distances from one of the graph vertices to all other ones and wrote them out in array d. Thus, element d[i] of the array shows the shortest distance from the vertex Valera chose to vertex number i.

Then something irreparable terrible happened. Valera lost the initial graph. However, he still has the array d. Help him restore the lost graph.

Input
The first line contains two space-separated integers n and k (1 ≤ k < n ≤ 105). Number n shows the number of vertices in the original graph. Number k shows that at most k edges were adjacent to each vertex in the original graph.

The second line contains space-separated integers d[1], d[2], …, d[n] (0 ≤ d[i] < n). Number d[i] shows the shortest distance from the vertex Valera chose to the vertex number i.

Output
If Valera made a mistake in his notes and the required graph doesn’t exist, print in the first line number -1. Otherwise, in the first line print integer m (0 ≤ m ≤ 106) — the number of edges in the found graph.

In each of the next m lines print two space-separated integers ai and bi (1 ≤ ai, bi ≤ n; ai ≠ bi), denoting the edge that connects vertices with numbers ai and bi. The graph shouldn’t contain self-loops and multiple edges. If there are multiple possible answers, print any of them.

Examples
Input
3 2
0 1 1
Output
3
1 2
1 3
3 2
Input
4 2
2 0 1 3
Output
3
1 3
1 4
2 3
Input
3 1
0 0 0
Output
-1

题意：有一幅$n$个点的无向无环图(那不就是树吗=_=)，边权全为$1$，现在只知道其中一个点到其他每个点的最短距离(与自己的距离是$0$)和每个点的最大度数$k$，请判断这样的图是否存在。存在则输出边数和每一条边，否则输出$-1$。

思路：把距离按升序排序，然后当做是一棵k叉树，一层一层连边。如果第$i$层所有的点的度数都到$k$了而第$i+1$层还有剩余的点，就输出$-1$
数据好坑啊，居然有一个点卡的是输入的距离没有$0$要输出$-1$

#include <cstdio>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;

const int maxn = 100005;
int n, k;
int degree[maxn];
vector<pair<int,int> > G;
struct node
{
    int val, id;

    bool operator<(const node &a) const
    {
        return val < a.val;
    }
} d[maxn];

int main()
{
    bool flag = 0;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; ++i)
    {
        scanf("%d", &d[i].val);
        d[i].id = i;
    }

    sort(d + 1, d + n + 1);

    if ((!d[1].val && !d[2].val) || d[1].val)//0的数量不为1时直接输出-1
    {
        printf("-1\n");
        return 0;
    }
    int fa = 1;
    for (int i = 2; i <= n; ++i)
    {
        while (d[i].val > d[fa].val + 1)
            ++fa;
        if (degree[d[fa].id] == k)
            ++fa;
        if (d[i].val != d[fa].val + 1)
        {
            flag = 1;
            break;
        }
        G.emplace_back(make_pair(d[fa].id,d[i].id));
        ++degree[d[fa].id];
        ++degree[d[i].id];
    }
    if (flag)
        printf("-1\n");
    else
    {
        printf("%d\n",(int)G.size());
        for (auto i:G)
            printf("%d %d\n",i.first,i.second);
    }
    return 0;
}