CF-27D. Ring Road 2（涂色）_d - ring road csdn-优快云博客

本文链接：https://blog.youkuaiyun.com/nealgavin/article/details/8576943

探讨了在银环城市间增加新道路的同时避免交叉的方法。通过分析道路连接关系，利用DFS深度优先搜索确定每条新道路的位置布局，确保道路系统有效且不发生交叉冲突。

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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";
  }
}