【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());
        }
    }
}
posted @ 2019-03-19 23:23  AWCXV  阅读(83)  评论(0编辑  收藏  举报