ICPC XIV NEERC Western Subregional Contest I题
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没有开头语
题目大意
给定一个无向图,构造子图使得所有节点的度数都为奇数
思路
对于每个连通块,构造一个生成树,如果一条边连着的子节点对应的子树大小为奇数,那这条边取,否则不取,复杂度为O(n+m)
代码
#include <bits/stdc++.h>using namespace std;typedef pair<int,int>pii;vector<int>g[100005];vector<pii>ans;bool vis[100005];int dfs(int u){ int cnt=0; vis[u]=true; for(int i=0;i<g[u].size();i++) { if(!vis[g[u][i]]&&dfs(g[u][i])%2==0) { ans.push_back(pii(u,g[u][i])); cnt++; } } return cnt;}int main(){ freopen("oddfactor.in","r",stdin); freopen("oddfactor.out","w",stdout); int n,m; memset(vis,0,sizeof vis); scanf("%d%d",&n,&m); for(int i=0;i<m;i++) { int a,b; scanf("%d%d",&a,&b); g[a].push_back(b); g[b].push_back(a); } bool f=true; for(int i=1;i<=n;i++) { if(!vis[i]&&dfs(i)%2==0) f=false; } if(f) { printf("%d\n",ans.size()); for(int i=0;i<ans.size();i++) { printf("%d %d\n",ans[i].first,ans[i].second); } }else{ printf("-1\n"); } return 0;}
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