B. Leha and another game about graph
time limit per test3 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Leha plays a computer game, where is on each level is given a connected graph with n vertices and m edges. Graph can contain multiple edges, but can not contain self loops. Each vertex has an integer di, which can be equal to 0, 1 or - 1. To pass the level, he needs to find a «good» subset of edges of the graph or say, that it doesn’t exist. Subset is called «good», if by by leaving only edges from this subset in the original graph, we obtain the following: for every vertex i, di = - 1 or it’s degree modulo 2 is equal to di. Leha wants to pass the game as soon as possible and ask you to help him. In case of multiple correct answers, print any of them.
Input
The first line contains two integers n, m (1 ≤ n ≤ 3·105, n - 1 ≤ m ≤ 3·105) — number of vertices and edges.
The second line contains n integers d1, d2, …, dn ( - 1 ≤ di ≤ 1) — numbers on the vertices.
Each of the next m lines contains two integers u and v (1 ≤ u, v ≤ n) — edges. It’s guaranteed, that graph in the input is connected.
Output
Print - 1 in a single line, if solution doesn’t exist. Otherwise in the first line k — number of edges in a subset. In the next k lines indexes of edges. Edges are numerated in order as they are given in the input, starting from 1.
Examples
input
1 0
1
output
-1
input
4 5
0 0 0 -1
1 2
2 3
3 4
1 4
2 4
output
0
input
2 1
1 1
1 2
output
1
1
input
3 3
0 -1 1
1 2
2 3
1 3
output
1
2
Note
In the first sample we have single vertex without edges. It’s degree is 0 and we can not get 1.
这个题的思路比较好，dfs，根据其条件根据已经走过的边的条件判断，具体看注释
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<iomanip>
//#define mem(dp,a) memset(dp,a,sizeof(dp))
//#define fo(i,n) for(int i=0;i<(n);i++)
//#define INF 0x3f3f3f3f
#define fread() freopen("data.txt","r",stdin)
#define fwrite() freopen("out.out","w",stdout)
using namespace std;
typedef long long ll;
const ll N = 300005;
const ll MOD = 1000000007;
int n,m;
int d[N];
int ed_in[N];
vector< pair<int,int> > e[N];
vector<int> ans;
bool vis[N];
void dfs(ll x,ll pre=0,ll e_id=0)
{
vis[x] = 1;
bool deg = 0;//判断连边的奇偶
for(auto u:e[x])
{
if(vis[u.first]) continue;
dfs(u.first,x,u.second);
if(ed_in[u.second] == 1)
{
deg = !deg;
}
}
//根据后面的状态来转移
if(d[x] == -1 || deg == d[x]) return ;
//如果不满足条件就加一条边
ed_in[e_id] = 1;
if(e_id!=0)
ans.push_back(e_id);
}
int main()
{
int u,v;
int sum = 0,di = -1;
ios_base::sync_with_stdio(false);
// fread();
cin >> n>> m;
for(int i=1;i<=n;i++)
{
cin >> d[i];
sum += d[i];
if(d[i] == -1)
di = i;
}
for(int i=1;i<=m;i++)
{
cin >> u >> v;
e[u].push_back({v,i});
e[v].push_back({u,i});
}
if(di!=-1) dfs(di);
else if(sum%2) return cout<<"-1\n",0;
else dfs(1);
cout <<ans.size()<<endl;
sort(ans.begin(),ans.end());
for(auto x:ans)
{
cout <<x<<endl;
}
return 0;
}