【Codeforces 404C】Restore Graph

【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());
        }
    }
}

原文地址：https://www.cnblogs.com/AWCXV/p/10562232.html