【链接】 我是链接,点我呀:)
【题意】
每个节点的度数不超过k
让你重构一个图
使得这个图满足 从某个点开始到其他点的最短路满足输入的要求
【题解】
把点按照dep的值分类
显然只能由dep到dep+1连边
设cnt[dep]表示到起点的距离为dep的点的集合
如果cnt[dep].size>cnt[dep+1].size
那么只要把dep层的前cnt[dep+1].size个点和dep+1层的点连就好了
否则
只能让dep层的点每个多连几个dep+1层的点了
【代码】
import java.io.*;
import java.util.*;
public class Main {
static InputReader in;
static PrintWriter out;
public static void main(String[] args) throws IOException{
//InputStream ins = new FileInputStream("E:\\rush.txt");
InputStream ins = System.in;
in = new InputReader(ins);
out = new PrintWriter(System.out);
//code start from here
new Task().solve(in, out);
out.close();
}
static int N = (int)1e5;
static class Task{
class Pair{
int x,y;
public Pair(int x,int y) {
this.x = x;
this.y = y;
}
}
int n,k;
ArrayList<Integer> g[] = new ArrayList[N+10];
ArrayList<Pair> ans = new ArrayList<>();
public void solve(InputReader in,PrintWriter out) {
for (int i = 0;i <= N;i++) g[i]=new ArrayList<Integer>();
n = in.nextInt();k = in.nextInt();
for (int i = 1;i <= n;i++) {
int d;
d = in.nextInt();
g[d].add(i);
}
if ((int)g[0].size()>1) {
out.println(-1);
}else {
for (int i = 1;i <= n;i++)
if ((int)g[i].size()>0 && g[i-1].isEmpty()) {
out.println(-1);
return;
}
for (int i = 1;i <= n;i++) {
if (g[i].isEmpty()) break;
int x = g[i].size();
int y = g[i-1].size();
//out.println(x+" "+y);
if (y>=x) {
for (int j = 0;j < x;j++) {
ans.add(new Pair(g[i-1].get(j),g[i].get(j)));
}
int ma = 1;
if (i-1>=1) {
ma = 2;
}
if (ma>k) {
out.println(-1);
return;
}
}else {
//y<x
int need = x/y;
if (x%y!=0) need++;
int ma = need;
//out.println(ma);
if (i-1>=1) ma++;
if (ma>k) {
out.println(-1);
return;
}
for (int j = 0;j < x;j++) {
int now = j;
now = now % y;
ans.add(new Pair(g[i-1].get(now),g[i].get(j)));
}
}
}
out.println((int)ans.size());
for (int i = 0;i < ans.size();i++) {
out.println(ans.get(i).x+" "+ans.get(i).y);
}
}
}
}
static class InputReader{
public BufferedReader br;
public StringTokenizer tokenizer;
public InputReader(InputStream ins) {
br = new BufferedReader(new InputStreamReader(ins));
tokenizer = null;
}
public String next(){
while (tokenizer==null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(br.readLine());
}catch(IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}