【CodeForces 990D】Graph And Its Complement

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luogu & CodeForces

题目描述

Given three numbers $\mathrm{n,a,bn,a,b.}$

$\mathrm{You\; need\; to\; find\; an\; adjacency\; matrix\; of\; such\; an\; undirected\; graph\; that\; the\; number\; of\; components\; in\; it\; is\; equal\; to aa,\; and\; the\; number\; of\; components\; in\; its\; complement\; is bb.\; The\; matrix\; must\; be\; symmetric,\; and\; all\; digits\; on\; the\; main\; diagonal\; must\; be\; zeroes.}$

In an undirected graph loops (edges from a vertex to itself) are not allowed. It can be at most one edge between a pair of vertices.

The adjacency matrix of an undirected graph is a square matrix of size $\mathrm{nn consisting\; only\; of\; "0"\; and\; "1",\; where nn is\; the\; number\; of\; vertices\; of\; the\; graph\; and\; the ii-th\; row\; and\; the ii-th\; column\; correspond\; to\; the ii-th\; vertex\; of\; the\; graph.\; The\; cell (i,j)(i,j) of\; the\; adjacency\; matrix\; contains 11 if\; and\; only\; if\; the ii-th\; and jj-th\; vertices\; in\; the\; graph\; are\; connected\; by\; an\; edge.}$

A connected component is a set of vertices $\mathrm{XX such\; that\; for\; every\; two\; vertices\; from\; this\; set\; there\; exists\; at\; least\; one\; path\; in\; the\; graph\; connecting\; this\; pair\; of\; vertices,\; but\; adding\; any\; other\; vertex\; to XX violates\; this\; rule.}$

The complement or inverse of a graph $\mathrm{GG is\; a\; graph HH on\; the\; same\; vertices\; such\; that\; two\; distinct\; vertices\; of HH are\; adjacent\; if\; and\; only\; if\; they\; are\; not\; adjacent\; in GG.}$

 

 题目翻译（摘自luogu）

一张无向图一共有n个点,a个联通块,它的补图一共有b个联通块,如果能构造出这张图输出'YES'并输出它的邻接矩阵，否则输出'NO'

输入:

三个整数n,a,b

输出：

如果有解输出'YES'和邻接矩阵，否则输出'NO'

感谢@zhaotiensn 提供翻译

解题思路

首先我们得要判断能否根据题目所给的数据构造出一个图，那怎么办呢？

代码

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int n,a,b;
 7 bool map[1005][1005];
 8 int main(){
 9     cin>>n>>a>>b;
10     if(a>1&&b>1){
11         cout<<"NO"<<endl;
12     }
13     else if(a>n||b>n){
14         cout<<"NO"<<endl;
15     }
16     else if(a==1&&b==1){
17         if(n<=3&&n>1)cout<<"NO"<<endl;
18         else {
19             cout<<"YES"<<endl;
20             for(register int i=1;i<n;i++){
21                 map[i][i+1]=map[i+1][i]=1;
22             }
23             for(register int i=1;i<=n;i++){
24                 for(register int j=1;j<=n;j++){
25                     printf("%d",map[i][j]);
26                 }
27                 putchar('\n');
28             }
29         }
30     }
31     else if(a>1){
32         cout<<"YES"<<endl;
33         for(register int i=a;i<=n;i++){
34             for(register int j=a;j<=n;j++){
35                 if(i==j)continue;
36                 map[i][j]=true;
37             }
38         }
39         for(register int i=1;i<=n;i++){
40             for(register int j=1;j<=n;j++){
41                 if(i==j)putchar('0');
42                 else printf("%d",map[i][j]);
43             }
44             putchar('\n');
45         }
46     }
47     else if(a==1){
48         cout<<"YES"<<endl;
49         for(register int i=b;i<=n;i++){
50             for(register int j=b;j<=n;j++){
51                 if(i==j)continue;
52                 map[i][j]=true;
53             }
54         }
55         for(register int i=1;i<=n;i++){
56             for(register int j=1;j<=n;j++){
57                 if(i==j)putchar('0');
58                 else printf("%d",1-map[i][j]);
59             }
60             putchar('\n');
61         }
62     }
63 }