nyoj-42

思路：(1).用搜索或者并查集判断是否连通 (2).用欧拉图判断是否可以一笔走完

并查集

#include<iostream>#include<cstdio>#include<string.h>using namespace std;int father[2222];int rank[2222];int degree[2222];int find(int x){return x==father[x]?x:find(father[x]);}void Union(int x,int y){x=find(x);y=find(y);if(x==y)return ;if(rank[x]>rank[y]){father[y]=x;rank[x]++;}else {if(rank[x]==rank[y])rank[y]++;father[x]=y;}}int main(){int i,t;cin>>t;while(t--){memset(degree,0,sizeof(degree));int n,m;cin>>n>>m;for(i=0;i<=n;++i)father[i]=i,rank[i]=0;int sum=0,cnt=0;while(m--){int x,y;cin>>x>>y;degree[x]++;degree[y]++;//x=find(x); 可以去掉这三个，也可以保留//y=find(y);//if(x!=y)Union(x,y);}for(i=1;i<=n;++i){if(find(i)==i)cnt++;if(degree[i]%2==1)sum++;}if((sum==0 || sum==2) && cnt==1)cout<<"Yes"<<endl;elsecout<<"No"<<endl;}return 0;}

　　深搜

#include<iostream>#include<cstdio>#include<string.h>using namespace std;int map[2222][2222],v[2222],degree[2222],P,Q;void dfs(int cur){int i;v[cur]=1;for(i=1;i<=P;++i)if(map[cur][i]){degree[cur]++;if(!v[i])dfs(i);}}int main(){int i,j,N;cin>>N;while(N--){int A,B,ok=1;memset(map,0,sizeof(map));memset(v,0,sizeof(v));memset(degree,0,sizeof(degree));cin>>P>>Q;for(i=0;i<Q;++i){cin>>A>>B;map[A][B]=1;map[B][A]=1;}dfs(1);for(i=1;i<=P;++i)if(!v[i]){ok=0;break;}j=0;for(i=1;i<=P;++i)if(degree[i]%2)j+=1;if((j==0 || j==2) && ok)cout<<"Yes"<<endl;elsecout<<"No"<<endl;}return 0;}