Hdu 5302 Connect the Graph 2015 Multi-University Training Contest 2

Connect the Graph

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 108    Accepted Submission(s): 36
Special Judge

Problem Description

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.

 

Input

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.

 

Output

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.

 

Sample Input

2 1 1 1 1 1 1 1 2 2 1 2 2

 

Sample Output

-1 6 1 5 0 4 5 0 2 4 0 1 4 1 1 3 1 2 3 1

 

Source

2015 Multi-University Training Contest 2

 

題目意思：

要求構造一組圖的解，滿足 白色邊：度爲0的點個數爲a[0],度爲1的點個數爲a[1]，度爲2的點的個數爲a[2]。

                                                                黑色邊：度爲0的點個數爲b[0],度爲1的點個數爲b[1]，度爲2的點的個數爲b[2]。

輸入a[0],a[1],a[2],b[0],b[1],b[2] 。且滿足sigma(a[i]) = sigma(b[i])

思路：總共點數爲sigma(a[i])。若a[1]&1 或b[1]&1 則無解。度爲1的點必須成對存在，在這隻有0 1 2度數的圖中。

清楚白色邊和黑色邊不相互影響，他們是獨立的。

總的邊數爲總度數/2 。即：（a[1]+a[2]*2）/2.

先構造白邊度數爲2的點，必然是（1,2）（2,3）（3,4）....(a[2]+1,a[2]+2)..如此下去，直到完成度數2的節點的構造。

再夠着白邊度數爲1的點，(a[2]+3,a[2]+4) (a[2]+5,a[2]+6).......

最後白色邊度數爲0的點，（,）(,)...(a[0]+a[1]+a[2],a[0]+a[1]+a[2])

黑色邊的處理需要間隔2的跳變（圖中兩點之間只有一條邊），其餘和白色邊構造類似。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define sf scanf
int T;
int a[5],b[5];
int d[1000005];
int main()
{
    //freopen("c.in","r",stdin);
    sf("%d",&T);
    while(T--)
    {
        for(int i = 0;i<3;i++)sf("%d",a+i);
        for(int i = 0;i<3;i++)sf("%d",b+i);
        int sum = 0;
        for(int i = 0;i<3;i++)sum+=a[i];
        if( (a[1]&1) || (b[1]&1) )
        {
            puts("-1");
            continue;
        }
        if(sum==4)
        {
            puts("4\n1 2 0\n1 3 0\n2 3 1\n3 4 1");
            continue;
        }
        printf("%d\n",a[1]/2+a[2]+b[1]/2+b[2]);
        int t = 1;
        while(a[2]!=-1){printf("%d %d 0\n",t,t+1);t++;a[2]--;}t++;
        while(a[1]!=2){printf("%d %d 0\n",t,t+1);t+=2;a[1]-=2;}

            int tt = 0;
            for(int i = 1;i<=sum;i+=2) d[tt++] = i;
            for(int i = 2;i<=sum;i+=2)d[tt++] = i;

        t = 0;
        while(b[2]!=-1){printf("%d %d 1\n",min(d[t],d[t+1]),max(d[t],d[t+1]));t++;b[2]--;}t++;
        while(b[1]!=2){printf("%d %d 1\n",min(d[t],d[t+1]),max(d[t],d[t+1]));t+=2;b[1]-=2;}

    }
}