CodeForces 990D Graph And Its Complement (FIGS complement and FIG configuration)

http://codeforces.com/problemset/problem/990/D

 

 

 

Meaning of the questions:

A configuration of order n simple undirected graph G, such that the number of communication branches to a, and the number of connected components of its complement graph b.

 

answer:

At first glance the title, and a look ignorant, ah? It makes me how to build map? ? ?

Or vegetables ah, look at someone else's problem solution learn from it. . .

Reference to: https://www.cnblogs.com/siuginhung/p/9172602.html

This is a configuration problem.

For an n-order simple undirected graph , if this does not communicate FIG G, it is the complement of FIG communication.

prove:

First, in a simple non-directed graph , if the node G in U, V ( U ≠ V) is not connected, then its complement figures, U, V bound communication.

FIG G = <V, E> is divided into k-branch communication, Gi of = <Vi, Ei>, I = 1,2, ..., k. In V to take any two points U, V ( U ≠ V).

若u∈Vi，v∈Vj，且i≠j，则u、v在图G中不连通，则u、v必然在其补图中连通；

若u,v∈Vi，则必然存在w∈Vj，且i≠j，使得u、w和v、w在补图中连通。

于是，在题中，a、b中至少有一个为1。

接下来构造连通分支：若一个n阶简单无向图有k（k≥2）个连通分支，则可以构造其连通分支分别为{1},{2},...,{k-1},{k,k+1,...,n}。

 

代码如下：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <string>
 5 #include <math.h>
 6 #include <algorithm>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <set>
11 #include <map>
12 #include <sstream>
13 const int INF=0x3f3f3f3f;
14 typedef long long LL;
15 #define Bug cout<<"---------------------"<<endl
16 const int mod=1e9+7;
17 const int maxn=2e5+10;
18 using namespace std;
19 
20 int G[1005][1005];
21 
22 int main()
23 {
24     int n,a,b;
25     scanf("%d %d %d",&n,&a,&b);
26     if((a!=1&&b!=1)||(n==2||n==3)&&(a==1&&b==1))
27         printf("NO\n");
28     else
29     {
30         printf("YES\n");
31         if(a==1)
32         {
33             for(int i=1;i<=n;i++)
34             {
35                 G[i][i]=0;
36                 for(int j=i+1;j<=n;j++)
37                     G[i][j]=G[j][i]=1;
38             }
39             for(int i=b;i<n;i++)
40                 G[i][i+1]=G[i+1][i]=0;
41         }
42         else
43         {
44             memset(G,0,sizeof(G));
45             for(int i=a;i<n;i++)
46             {
47                 G[i][i+1]=G[i+1][i]=1;
48             }
49         }
50         for(int i=1;i<=n;i++)
51         {
52             for(int j=1;j<=n;j++)
53                 printf("%d",G[i][j]);
54             printf("\n");
55         }
56     }
57     return 0;
58 }