NYOJ - 1015 二部圖(bfs/dfs)
阿新 • • 發佈:2017-09-15
for esp 題意 const ring cto style ios line
題目鏈接:點我點我
題意:二分圖判斷問題
題解:兩種解法,模擬下匹配過程。
1 //二分圖匹配dfs 2 #include <cstring> 3 #include <vector> 4 #include <iostream> 5 #include <algorithm> 6 using namespace std; 7 8 const int N=11111; 9 vector <int> E[N]; 10 int col[N]; 11 12 void init(){ 13 memset(col,0,sizeof(col)); 14 for(int i=0;i<N;i++) E[i].clear(); 15 } 16 17 bool dfs(int v,int c){ 18 col[v]=c; 19 for(int i=0;i<E[v].size();i++){ 20 int tmp=E[v][i]; 21 if(col[tmp]==c) return false; 22 if(col[tmp]==0&&!dfs(tmp,-c)) return false; 23 }24 return true; 25 } 26 27 int main(){ 28 int n,m,x,y; 29 while(cin>>n){ 30 int idx=0; 31 init(); 32 cin>>m; 33 for(int i=1;i<=m;i++){ 34 cin>>x>>y; 35 E[x].push_back(y); 36 E[y].push_back(x);37 } 38 if(!dfs(0,1)) cout<<"NOT BICOLORABLE."<<endl; 39 else cout<<"BICOLORABLE."<<endl; 40 } 41 return 0; 42 }
1 //二分圖匹配bfs 2 #include <queue> 3 #include <cstring> 4 #include <vector> 5 #include <iostream> 6 #include <algorithm> 7 using namespace std; 8 9 const int N=11111; 10 vector <int> E[N]; 11 int col[N]; 12 13 void init(){ 14 memset(col,0,sizeof(col)); 15 for(int i=0;i<N;i++) E[i].clear(); 16 } 17 18 bool bfs(int x){ 19 col[x]=1; 20 queue <int> Q; 21 Q.push(x); 22 while(!Q.empty()){ 23 int now=Q.front();Q.pop(); 24 for(int i=0;i<E[now].size();i++){ 25 int tmp=E[now][i]; 26 int ctmp=col[now]==1?2:1; 27 if(col[tmp]==0){ 28 col[tmp]=ctmp; 29 Q.push(tmp); 30 } 31 else if(col[tmp]!=ctmp) return false; 32 } 33 } 34 return true; 35 } 36 37 int main(){ 38 int n,m; 39 while(cin>>n){ 40 init(); 41 int x,y; 42 cin>>m; 43 for(int i=1;i<=m;i++){ 44 cin>>x>>y; 45 E[x].push_back(y); 46 E[y].push_back(x); 47 } 48 if(bfs(0)) cout<<"BICOLORABLE."<<endl; 49 else cout<<"NOT BICOLORABLE."<<endl; 50 } 51 return 0; 52 }
NYOJ - 1015 二部圖(bfs/dfs)