POJ 1659 Frogs' Neighborhood havel度序列定理的应用

havel度序列定理就是个贪心，每次先选度数大的点和其他点连一条边

证明画个图就可以了，如果度数最大的那个点没有和度数靠前的点连边的话那么一定可以通过交换边的方式让它和所有度数靠前的点连边，那么就能证明havel定理的转换过程是充要的~

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
using namespace std;

#ifdef _WIN32
#define i64 __int64
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
#define i64 long long
#define out64 "%lld\n"
#define in64 "%lld"
#endif

#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a)        for( int i = (a)-1 ; i >= 0 ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define SZ(a)           ((int)a.size())
#define PP(n,m,a)		puts("---");FF(i,n){FF(j,m)cout << a[i][j] << ' ';puts("");}
#define pb              push_back
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const i64 inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-10;
const double pi = acos(-1.0);
const int maxn = 13;

struct zz
{
    int id;
    int d;
    bool operator < (const zz & cmp ) const
    {
        return d < cmp.d;
    }
}zx[maxn];
int n,T;
int a[maxn][maxn];

void make(int x)
{
    sort(zx+x,zx+n+1);
    reverse(zx+x,zx+n+1);
    return ;
}

bool can()
{
    memset(a,0,sizeof(a));
    int from,to;
    for(int k=1;k<=n;k++)
    {
        make(k);
        from = zx[k].id;
        for(int i=k+1;i<=n;i++)
        {
            to = zx[i].id;
            if(zx[k].d)
            {
                zx[k].d--;
                if(zx[i].d)
                {
                    zx[i].d--;
                    a[from][to]++;
                    a[to][from]++;
                }
                else
                {
                    return false;
                }
            }
            else
            {
                break;
            }
        }
        if(zx[k].d)
        {
            return false;
        }
    }
    return true;
}

int main()
{
    SS(T);
    while(T--)
    {
        SS(n);
        FOR(i,1,n)
        {
            zx[i].id = i;
            SS(zx[i].d);
        }
        if(can())
        {
            cout<<"YES"<<endl;
            FOR(i,1,n)
            {
                FOR(j,1,n-1)
                {
                    cout<<a[i][j]<<" ";
                }
                cout<<a[i]
<<endl;
            }
        }
        else
        {
            cout<<"NO"<<endl;
        }
        cout<<endl;
    }
    return 0;
}