2015
09-17

Assignment For Princess

Long long ago, in the Kingdom Far Far Away, there lived many little animals. And you are the beloved princess who is marrying the prince of a rich neighboring kingdom. The prince, who turns out to be a handsome guy, offered you a golden engagement ring that can run computer programs!
The wedding will be held next summer because your father, the king, wants you to finish your university first.
But you did’t even have a clue on your graduation project. Your terrible project was to construct a map for your kingdom. Your mother, the queen, wanted to make sure that you could graduate in time.
Or your wedding would have to be delayed to the next winter. So she told you how your ancestors built the kingdom which is called the Roads Principle:

2. There is at most one road between a pair of castles.
3. There won’t be any roads that start at one castle and lead to the same one.
1. No matter which castle you start with, you can arrive at any other castles.
2. Traveling on theM roads will take 1, 2, 3, … ,M days respectively, no two roads need the same number of days.
3. You can take a round trip starting at any castle, visiting a sequence of castles, perhaps visiting some castles or traveling on some roads more than once, and finish your journey where you started.
4. The total amount of days spent on any round trip will be a multiple of three.
But after a month, you still couldn’t make any progress. So your brother, the future king, asked your university to assign you a simpler project. And here comes the new requirements. Construct a map that satisfies both the Roads Principle and the Travelers Guide when N and M is given.
There would probably be several solutions, but your project would be accepted as long as it meets the two requirements.
Perhaps they could finish it within 5 hours and you can think of your sweet wedding now.

The first line contains only one integer T, which indicates the number of test cases.
For each test case, there is one line containing two integers N, M described above.(10 <= N <= 80, N+3 <= M <= N2/7 )

The first line contains only one integer T, which indicates the number of test cases.
For each test case, there is one line containing two integers N, M described above.(10 <= N <= 80, N+3 <= M <= N2/7 )

1
6 8

Case #1:
1 2 1
2 3 2
2 4 3
3 4 4
4 5 5
5 6 7
5 1 6
6 1 8
Hint
��The restrictions like N >= 10 will be too big for a sample. So the sample is just a simple case for the detailed formats of input and output,
and it may be helpful for a better understanding. Anyway it won’t appear in actual test cases.


1.强连通。

2.任意俩点直接之间只有一条有向边，自己和自己无边。

3.任意一个闭合回路权和%3为0。

4.每条边的权理论不同，而且是1,2,3..m

：注意若 在 i—>j 附加边，则有w（i,j）%3==从i到j边权之和%3即可（容易证明）

#include<iostream>
#include<cstring>
using namespace std;
int n,m;
int mark[6400];
int vis[85][85];
int main()
{
int T;
cin>>T;int ct=1;
while(T--)
{
for(int i=0;i<=m;i++)
mark[i]=0;
memset(vis,0,sizeof(vis));
cin>>n>>m;
for(int i=1;i<n;i++)
{
// cout<<i<<" "<<i+1<<" "<<i<<endl;
mark[i]=1;vis[i][i+1]=vis[i+1][i]=i;
}
int tei=0;
for(int i=n;i<n+3;i++)
if((n*(n-1)/2+i)%3==0)
{
//cout<<n<<" "<<1<<" "<<i<<endl;
tei=i;
mark[i]=1;
vis[n][1]=vis[1][n]=i;
break;
}
for(int i=n;i<=m;i++)
{
if(!mark[i])
{
int flags=0;
for(int j=1;j<=n;j++)
{

for(int k=j+2;k<=n;k++)
{

if(!vis[j][k]&&((k-j)*(j+k-1)/2)%3==i%3)
{
vis[j][k]=vis[k][j]=i;
mark[i]=1;
flags=1;
break;
}
}
if(flags)break;
}
}
}
bool flag=1;
for(int i=1;i<=m;i++)
{
if(mark[i]==0)
{
flag=0;break;
}
}
cout<<"Case #"<<ct++<<":"<<endl;
if(flag==0)
cout<<-1<<endl;
else
{
for(int i=1;i<=n;i++)
for(int j=(i+1);j<=n;j++)
{
if(vis[i][j]!=0&&!(i==1&&j==n))
{
cout<<i<<" "<<j<<" "<<vis[i][j]<<endl;
}
}
cout<<n<<" "<<1<<" "<<tei<<endl;
}
}
}