题面:
You are given a connected undirected graph consisting of nn vertices and mm edges. There are no self-loops or multiple edges in the given graph.
You have to direct its edges in such a way that the obtained directed graph does not contain any paths of length two or greater (where the length of path is denoted as the number of traversed edges).
The first line contains two integer numbers nn and mm (2≤n≤2⋅1052≤n≤2⋅105, n−1≤m≤2⋅105n−1≤m≤2⋅105) — the number of vertices and edges, respectively.
The following mm lines contain edges: edge ii is given as a pair of vertices uiui, vivi (1≤ui,vi≤n1≤ui,vi≤n, ui≠viui≠vi). There are no multiple edges in the given graph, i. e. for each pair (ui,viui,vi) there are no other pairs (ui,viui,vi) and (vi,uivi,ui) in the list of edges. It is also guaranteed that the given graph is connected (there is a path between any pair of vertex in the given graph).
If it is impossible to direct edges of the given graph in such a way that the obtained directed graph does not contain paths of length at least two, print "NO" in the first line.
Otherwise print "YES" in the first line, and then print any suitable orientation of edges: a binary string (the string consisting only of '0' and '1') of length mm. The ii-th element of this string should be '0' if the ii-th edge of the graph should be directed from uiui to vivi, and '1' otherwise. Edges are numbered in the order they are given in the input.
6 5
1 5
2 1
1 4
3 1
6 1
YES
10100
题目大意:给定一个无向图,要求通过添加方向使其成为有向图,条件是不能存在长度大于1的路径。其中每个点所有的边只能有指向它或者背向它。
解题思路:二分匹配
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<cassert>
#include<complex>//real().imag().
#include<cctype>
#include<algorithm>
#include<iomanip>
#include<stack>
#include<queue>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<string>
#include<utility>
#include<iterator>
#define pii pair<int,int>
#define make_pair mp
using namespace std;
typedef long long ll;
const int N = 200005;
int n, m,u,v,flag=1;
vector<int> vc[N];
int color[N];
int solve[N];
void dfs(int u, int cl)
{
color[u] = cl;
for (int i = 0, j = vc[u].size(); i < j; i++)
{
int t = vc[u][i];
if (color[u] == color[t])
{
flag = 0;
}
if (color[t] == -1)
{
dfs(t, !cl);
}
}
}
int main()
{
// freopen("D:\\in.txt","r",stdin);
// freopen("D:\\out.txt","w",stdout);
//ios::sync_with_stdio(false);
cin >> n >> m;
for (int i = 0; i < m; i++)
{
cin >> u >> v;
vc[u].push_back(v);
vc[v].push_back(u);
solve[i] = u;
}
memset(color, -1, sizeof(color));
for (int i = 1; i <= n; ++i)
{
if (color[i] == -1)
{
dfs(i, 0);
}
}
if (flag)
{
cout << "YES" << endl;
for (int i = 0; i < m; i++)
{
cout << !color[solve[i]];
}
}
else
cout << "NO" << endl;
//system("pause");
return 0;
}