Connect the Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 456    Accepted Submission(s): 144

Special Judge

Once there was a special graph. This graph had
n
vertices and some edges. Each edge was either white or black. There was no edge connecting one vertex and the vertex itself. There was no two edges connecting the same pair of vertices. It is special because the each vertex is connected to at most two black
edges and at most two white edges.



One day, the demon broke this graph by copying all the vertices and in one copy of the graph, the demon only keeps all the black edges, and in the other copy of the graph, the demon keeps all the white edges. Now people only knows there are
w0
vertices which are connected with no white edges,
w1
vertices which are connected with 1
white edges, w2
vertices which are connected with 2
white edges, b0
vertices which are connected with no black edges,
b1
vertices which are connected with 1
black edges and b2
vertices which are connected with 2
black edges.



The precious graph should be fixed to guide people, so some people started to fix it. If multiple initial states satisfy the restriction described above, print any of them.

The first line of the input is a single integer
T (T≤700),
indicating the number of testcases.



Each of the following T
lines contains w0,w1,w2,b0,b1,b2.
It is guaranteed that 1≤w0,w1,w2,b0,b1,b2≤2000
and b0+b1+b2=w0+w1+w2.




It is also guaranteed that the sum of all the numbers in the input file is less than
300000.

For each testcase, if there is no available solution, print
−1.
Otherwise, print m
in the first line, indicating the total number of edges. Each of the next
m
lines contains three integers x,y,t,
which means there is an edge colored t
connecting vertices x
and y.
t=0
means this edge white, and t=1
means this edge is black. Please be aware that this graph has no self-loop and no multiple edges. Please make sure that
1≤x,y≤b0+b1+b2.

2

1 1 1 1 1 1

1 2 2 1 2 2

-1

6

1 5 0

4 5 0

2 4 0

1 4 1

1 3 1

2 3 1

构造法：

首先保证度数之和为偶数，即w1=b1=1 ,否则无解

又w0,w1,w2,b0,b1,b2均为正数 故

当n=4时,仅仅有1种情况 1 2 1 不是无解

当n≥4时,先构造2个环分别为白环，黑环

对于奇数n：



　　白环 1 2 3 ... n



　　黑环 1 3 5 ... n 2 4 6 ... n-1



对于偶数n：



　　白环 1 2 3 ... n



　　黑环 1 3 5 ... n-1 2 n n-2 n-4 ... 4

此时，对于每一个环而言，构造答案

1-2-2-...-2-2-1
1-1 1-1 .. 1-1
1-1 0 .. 0

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<algorithm>

#include<functional>

#include<iostream>

#include<cmath>

#include<cctype>

#include<ctime>

using namespace std;

#define For(i,n) for(int i=1;i<=n;i++)

#define Fork(i,k,n) for(int i=k;i<=n;i++)

#define Rep(i,n) for(int i=0;i<n;i++)

#define ForD(i,n) for(int i=n;i;i--)

#define RepD(i,n) for(int i=n;i>=0;i--)

#define Forp(x) for(int p=pre[x];p;p=next[p])

#define Forpiter(x) for(int &p=iter[x];p;p=next[p])

#define Lson (x<<1)

#define Rson ((x<<1)+1)

#define MEM(a) memset(a,0,sizeof(a));

#define MEMI(a) memset(a,127,sizeof(a));

#define MEMi(a) memset(a,128,sizeof(a));

#define INF (2139062143)

#define F (100000007)

#define MAXD (2000+10)

#define MAXN (6000+10)

typedef long long ll;

ll mul(ll a,ll b){return (a*b)%F;}

ll add(ll a,ll b){return (a+b)%F;}

ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}

void upd(ll &a,ll b){a=(a%F+b%F)%F;}

int a2[MAXN],a1[MAXN],n;

void calc(int *a,int n0,int n1,int n2,int p)

{

	int i=1;

	if (n1==0&&n2==0) return;

	For(i,n2+1)

	{

		printf("%d %d %d\n",a[i],a[i+1],p);

	}

	n1-=2;

	for(int i=n2+3,j=1;j<=n1;i+=2,j+=2) printf("%d %d %d\n",a[i],a[i+1],p);

}

int main()

{

//	freopen("C.in","r",stdin);

//	freopen(".out","w",stdout);

	int T; cin>>T;

	while(T--) {

		int w0,w1,w2,b0,b1,b2;

		scanf("%d%d%d%d%d%d",&w0,&w1,&w2,&b0,&b1,&b2);

		n=w0+w1+w2;

		//特判

		if ((w1&1)||(b1&1)) { printf("-1\n");continue;}

		int m=(w1+2*w2+b1+2*b2)/2;

		if (n==4)

		{

			puts("4\n1 2 0\n1 3 0\n2 3 1\n3 4 1");

			continue;

		}

		else if (n>4) {

			For(i,n) a1[i]=i;

			if (n%2==0)

			{

				for(int i=1,j=1;i<=n/2;i++,j+=2) a2[i]=j;

				for(int i=n/2+1,j=2;i<=n;i++,j+=2) a2[i]=j;

				a2[n+1]=1;

			}

			else {

				for(int i=1,j=1;i<=n/2+1;i++,j+=2) a2[i]=j;

				a2[n/2+2]=2;

				for(int i=n/2+3,j=n-1;i<=n;i++,j-=2) a2[i]=j;

				a2[n+1]=1;

			}

			cout<<m<<endl;

			calc(a1,w0,w1,w2,0);

			calc(a2,b0,b1,b2,1);

		}

	}

	return 0;

}