题意:构造一个图,使这个图的连通分量有a个,其补图的连通分量有b个,输出邻接矩阵
可以推出当min(a,b)!=1时输出no
a=b=1且n=2或者n=3时也为no
其余只要把一个连通分量里的x个点用x-1条边串起来就好了
哎,最后想到n=3也为no,可惜了..
#include <bits/stdc++.h>
#define ll long long
#define pb push_back
#define inf 0x3f3f3f3f
#define pll pair<ll,ll>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define rep1(i,a,b) for(int i=a;i>=b;i--)
#define rson rt<<1|1,m+1,r
#define lson rt<<1,l,m
#define mod 323232323
using namespace std;
const int N=1000+100;
int arr[N][N];int main()
{#ifdef LOCAL_DEFINEfreopen("D:\\rush.txt","r",stdin);#endifios::sync_with_stdio(false),cin.tie(0);int n,a,b;cin>>n>>a>>b;if((n==2||n==3)&&a==1&&b==1){cout<<"NO"<<endl;return 0;}if(min(a,b)!=1){cout<<"NO"<<endl;}else{cout<<"YES"<<endl;int c=max(a,b);for(int i=1;i<=n-c;i++){arr[i][i+1]=1;arr[i+1][i]=1;}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(i==j)cout<<0;else if(c!=a)cout<<!arr[i][j];elsecout<<arr[i][j];}cout<<endl;}}return 0;
}
