题目链接：点击打开链接

题意：

给定一个无向图

任选一个起点，使得访问每个点的次数奇偶与最后一行的输入一致

思路：

选一个奇数点作为起点

dfs树一下，若子节点不满足则先走到子节点再回来

如此就能保证所有子节点合法

这样dfs结束后最多只有根节点不满足，随便找一个与根相连的点调整一下。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <math.h>
#include <queue>
#include <set>
#include <algorithm>
#include <stdlib.h>
using namespace std;
#define ll int
#define N 100005
#define inf 10000009
vector<int>G[N],ans, one, tmp;
int n, m;
int a[N], st, dis[N], pre[N];
void go(int u, int v, vector<int>&x){// v - u 
    x.push_back(v);
    while(1){
        v = pre[v];
        x.push_back(v);
        if(v==u)return ;
    }
}
int vis[N], hehe[N];
void dfs(int u, int fa){
    hehe[u] = fa;
    ans.push_back(u);
    vis[u] ^= 1;
    for(int i = 0; i < G[u].size(); i++)
        if(hehe[G[u][i]]==0)
        {
            dfs(G[u][i], u);
            if(a[G[u][i]] != vis[G[u][i]])
            {
                ans.push_back(u);
                ans.push_back(G[u][i]); 
                vis[G[u][i]]^=1;
                vis[u] ^= 1;
            }

        if(ans[ans.size()-1] != u){ans.push_back(u); vis[u]^=1;}
        }
}

void input(){
    one.clear();
    for(int i = 0; i <= n; i++)G[i].clear();
    while(m--){
        int u, v;
        scanf("%d %d",&u,&v);
        G[u].push_back(v);      G[v].push_back(u);
    }
    st = -1;
    for(int i = 1; i <= n; i++){
        scanf("%d",&a[i]);
        if(a[i])
        {
            if(st==-1)st = i;
            one.push_back(i);
        }
    }
}
int main(){
    int i, j, u, v;
    while(~scanf("%d %d",&n,&m))
    {
        input();
        if(st == -1)  {puts("0");continue;}
        if(one.size()==1){printf("1\n%d\n", st);continue;}
        ans.clear();
        memset(vis,0,sizeof vis);
        memset(hehe, 0, sizeof hehe);
        hehe[st] = st;
        dfs(st, st);
        if(a[st]!=vis[st]){
            ans.push_back(G[st][0]);
            ans.push_back(st);
            ans.push_back(G[st][0]);
        }
        bool ok = false;
        for(i = 1; i <= n; i++)if(a[i] && hehe[i] == 0)ok = true;
        if(ok){puts("-1");continue;}

        printf("%d\n",ans.size());
        for(i = 0; i < ans.size(); i++)printf("%d ",ans[i]);puts("");
        //memset(a, 0, sizeof a);       for(i = 0; i < ans.size(); i++)a[ans[i]]^=1;        for(i = 1; i <=n ;i++)printf("%d ", a[i]);puts("");
    }
    return 0;
}