CF 990D Graph And Its Complement 第十八 构造、思维

Graph And Its Complement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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

Input

In a single line, three numbers are given $\mathrm{n,a,b(1\le n\le 1000,1\le a,b\le n)n,a,b(1\le n\le 1000,1\le a,b\le n):\; is\; the\; number\; of\; vertexes\; of\; the\; graph,\; the\; required\; number\; of\; connectivity\; components\; in\; it,\; and\; the\; required\; amount\; of\; the\; connectivity\; component\; in\; it\text{'}s\; complement.}$

Output

If there is no graph that satisfies these constraints on a single line, print "NO" (without quotes).

Otherwise, on the first line, print "YES"(without quotes). In each of the next $\mathrm{nn lines,\; output nn digits\; such\; that jj-th\; digit\; of ii-th\; line\; must\; be 11 if\; and\; only\; if\; there\; is\; an\; edge\; between\; vertices ii and jj in GG (and 00 otherwise).\; Note\; that\; the\; matrix\; must\; be\; symmetric,\; and\; all\; digits\; on\; the\; main\; diagonal\; must\; be\; zeroes.}$

If there are several matrices that satisfy the conditions — output any of them.

Examples

input

Copy

3 1 2

output

Copy

YES
001
001
110

input

Copy

3 3 3

output

Copy

NO

 引用别人的题解了。。。

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

这是一个构造问题。

构造一张n阶简单无向图G，使得其连通分支个数为a，且其补图的连通分支个数为b。

对于一张n阶简单无向图G，若此图不连通，则其补图是连通的。

证明：

首先，在简单无向图G中，若结点u、v（u≠v）不连通，则在其补图中，u、v必然连通。

将图G=<V,E>划分为k个连通分支，Gi=<Vi,Ei>，i=1,2,...,k。在V中任取两点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}。

这是我的代码

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e3 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int mapn[maxn][maxn];
int main(){
    std::ios::sync_with_stdio(false);
    ll n, a, b;
    while( cin >> n >> a >> b ) {
        bool flag = true;
        if( a != 1 && b != 1 ) {
            flag = false;
        }
        if( ( n == 2 || n == 3 ) && ( a + b == 2 ) ) {
            flag = false;
        }
        if( !flag ) {
            cout << "NO" << endl;
            continue;
        }
        cout << "YES" << endl;
        if( b == 1 ) {
            memset( mapn, 0, sizeof(mapn) );
            for( ll i = a; i < n; i ++ ) {
                 mapn[i-1][i] = 1;
                 mapn[i][i-1] = 1;
            }
        } else {
            memset( mapn, -1, sizeof(mapn) );
            for( ll i = 0; i < n; i ++ ) {
                mapn[i][i] = 0;
            }
            for( ll i = b; i < n; i ++ ) {
                mapn[i-1][i] = 0;
                mapn[i][i-1] = 0;
            }
        }
        for( ll i = 0; i < n; i ++ ) {
            for( ll j = 0; j < n; j ++ ) {
                putchar( mapn[i][j] ? '1' : '0' );
            }
            putchar('
');
        }
    }
    return 0;
}