CF-27D. Ring Road 2（涂色）

D. Ring Road 2

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It is well known that Berland has n cities, which form the Silver ring — cities i and i + 1 (1 ≤ i < n) are connected by a road, as well as the cities n and 1. The goverment have decided to build m new roads. The list of the roads to build was prepared. Each road will connect two cities. Each road should be a curve which lies inside or outside the ring. New roads will have no common points with the ring (except the endpoints of the road).

Now the designers of the constructing plan wonder if it is possible to build the roads in such a way that no two roads intersect (note that the roads may intersect at their endpoints). If it is possible to do, which roads should be inside the ring, and which should be outside?

Input

The first line contains two integers n and m (4 ≤ n ≤ 100, 1 ≤ m ≤ 100). Each of the following m lines contains two integers a_i and b_i(1 ≤ a_i, b_i ≤ n, a_i ≠ b_i). No two cities will be connected by more than one road in the list. The list will not contain the roads which exist in the Silver ring.

Output

If it is impossible to build the roads in such a way that no two roads intersect, output Impossible. Otherwise print m characters. i-th character should be i, if the road should be inside the ring, and o if the road should be outside the ring. If there are several solutions, output any of them.

Sample test(s)

input

4 2
1 3
2 4

output

io

input

6 3
1 3
3 5
5 1

output

ooo

思路：就三种情况，相交，一个在另一段里，在另一段外，每次选择一段相交的涂不同颜色，如果涂到已图色，且和需要图的颜色不同则无解

#include<iostream>
#include<cstring>
using namespace std;
const int mm=110;
int a[mm],b[mm],c[mm];
int n,m;
bool flag;
void dfs(int u,int col)
{
  if(c[u]==-1)
  {
    c[u]=col;
    for(int i=0;i<m;i++)///找相交
    {
      if(a[u]!=a[i]&&b[u]!=b[i])
      {
        if(a[u]<a[i]&&b[u]>a[i]&&b[u]<b[i])
          dfs(i,col^1);
        if(a[i]<a[u]&&b[i]>a[u]&&b[i]<b[u])
          dfs(i,col^1);
      }
    }
  }
  else if(c[u]^col)flag=0;
}
int main()
{
  while(cin>>n>>m)
  { flag=1;
    memset(c,-1,sizeof(c));
    for(int i=0;i<m;i++)
    {
      cin>>a[i]>>b[i];a[i]--,b[i]--;
      if(a[i]>b[i])a[i]^=b[i],b[i]^=a[i],a[i]^=b[i];///a<b
    }
    for(int i=0;i<m;i++)
      if(c[i]==-1)
      dfs(i,0);
    if(flag)
      for(int i=0;i<m;i++)
      cout<<(c[i]?'i':'o');
    else cout<<"Impossible\n";
    cout<<"\n";
  }
}