Codeforces 404C Restore Graph【思维】

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本文探讨了如何根据给定的最短路径信息恢复一个无向图的问题。该图包含N个节点，每个节点的度数不超过K，目标是从指定起始点出发到达其他各点的最短路径距离符合给定条件。

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, elementd[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

题目大意：

让你构造出来一个图，使得其包含N个点，并且保证每个点的度都不超过K，要求从起点到各个点的最短路为d【i】，问是否可以构造出可行解，如果可以，输出边数和边的信息，否则输出-1.

思路：

我们模拟跑最短路的过程即可：

①我们首先找到起点，然后处理最短路长度为1的各个点，我们将起点连入这些点，都连一条边

②然后我们处理长度为2的各个点，我们将这些个点，和最短路长度为1的某一个点连一条边即可，当一个点的度达到了K之后，这个点将不会再和其他点相连。

③然后我们处理长度为3的各个点，我们将这些个点，和最短路长度为2的某一个点连一条边即可，当一个点的度达到了K之后，这个点将不会再和其他点相连。

依次类推，我们只要能够将整个图建立出来就是包含可行解，否则就是-1.

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
struct node
{
    int x,y;
}ans[2050000];
int degree[150000];
queue<int>s[100002];
vector<int>mp[150000];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        int flag=0;
        memset(degree,0,sizeof(degree));
        for(int i=1;i<=n;i++)
        {
            while(!s[i].empty())s[i].pop();
            mp[i].clear();
        }
        for(int i=1;i<=n;i++)
        {
            int x;scanf("%d",&x);
            if(x==0)flag++;
            s[x].push(i);
            mp[x].push_back(i);
        }
        if(flag!=1)
        {
            printf("-1\n");
            continue;
        }
        int cnt=0;
        int ok=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=0;j<mp[i].size();j++)
            {
                int u=-1;
                int v=mp[i][j];
                while(!s[i-1].empty())
                {
                    u=s[i-1].front();
                    if(degree[u]<k)break;
                    else s[i-1].pop();
                }
                if(u!=-1&°ree[u]<k)
                {
                    degree[u]++;
                    degree[v]++;
                    ans[cnt].x=u;
                    ans[cnt++].y=v;
                }
                else ok=0;
            }
        }
        if(ok==1)
        {
            printf("%d\n",cnt);
            for(int i=0;i<cnt;i++)
            {
                printf("%d %d\n",ans[i].x,ans[i].y);
            }
        }
        else printf("-1\n");
    }
}