codeforces Round #237(div2) C解题报告

C. Restore Graph

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

input

3 20 1 1

output

31 21 33 2

input

4 22 0 1 3

output

31 31 42 3

input

3 10 0 0

output

-1

题目大意:

        给出从某一点到其他所有点的最短路径

        求出该图

解法:

        判断好-1,模拟即可

#include <cstdio>#include <vector>using namespace std;int n, k, len;vector<int> dis[1000010];vector<pair<int,int> > ans;void init() {scanf("%d%d",&n,&k);for (int i = 1; i <= n; i++) {int tmp;scanf("%d", &tmp);dis[tmp].push_back(i);len = max(len, tmp);}}void solve() {if (dis[0].size() != 1) {printf("-1");return;}for (int i = 1; i <= len; i++)  {int edge = (i != 1), total = 0;for (int j = 0; j < dis[i].size(); j++) {if (edge == k) {edge = (i != 1);total++;}if (total == dis[i-1].size()) {printf("-1\n");return;}ans.push_back(make_pair(dis[i-1][total], dis[i][j]));edge++;}}printf("%d\n", ans.size());for (int i = 0; i < ans.size(); i++)printf("%d %d\n", ans[i].first, ans[i].second);}int main() {init();solve();}