hdu 1325 is it a tree?
网上一搜全是并查集。。 并查集POJ可以水过,HDU就不行了。。
并查集的话:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int pre[1005]; 6 bool vis[1005]; 7 int num[1005]; 8 bool flag; 9 int x,y,icase=1; 10 void init() 11 { 12 for(int i=0;i<=1000;i++) 13 pre[i]=i; 14 memset(num,0,sizeof(num)); 15 memset(vis,false,sizeof(vis)); 16 flag=false; 17 } 18 int find(int x) 19 { 20 if(pre[x]!=x) 21 pre[x]=find(pre[x]); 22 return pre[x]; 23 } 24 void judge() 25 { 26 for(int i=1;i<=1000;i++) 27 if(vis[i]) num[find(i)]++; 28 int key=0; 29 for(int i=1;i<=1000;i++) 30 if(num[i]>1) key++; 31 if(key>1) flag=true; 32 } 33 int main() 34 { 35 // freopen("in.txt","r",stdin); 36 init(); 37 while(1) 38 { 39 scanf("%d%d",&x,&y); 40 if(x<0 && y<0) 41 break; 42 else if(x==0 && y==0) 43 { 44 judge(); 45 if(flag) 46 printf("Case %d is not a tree.\n",icase++); 47 else 48 printf("Case %d is a tree.\n",icase++); 49 init(); 50 continue; 51 } 52 else 53 { 54 vis[x]=vis[y]=true; 55 if(find(x)==find(y) || pre[y]!=y) 56 flag=true; 57 else 58 pre[y]=x; 59 } 60 } 61 return 0; 62 }
有个坑是只要是输入两个负数就结束,不一定是-1 -1
判断图是否为树:1.是否为连通图 2.是否有环
因为题目给的是从父亲指向儿子,所以直接判断点的入度就可以了
1 #include "iostream" 2 #include "memory.h" 3 #include "cstdio" 4 using namespace std; 5 #define MAXN 1111 6 int fa[MAXN]; 7 int in_degree[MAXN]; 8 bool c[MAXN]; 9 bool ok; 10 11 int main() 12 { 13 int a,b,T = 1; 14 while (~scanf("%d%d",&a,&b)) { 15 memset(fa,-1,sizeof(fa)); 16 memset(in_degree,0,sizeof(in_degree)); 17 memset(c,0,sizeof(c)); 18 if (a < 0 && b < 0) break; 19 c[a] = 1; 20 c[b] = 1; 21 in_degree[b]++; 22 while (~scanf("%d%d",&a,&b),a+b) { 23 if (a < 0 && b < 0) return 0; 24 in_degree[b]++; 25 c[a] = 1; 26 c[b] = 1; 27 } 28 int re = 0; 29 int ok = 1; 30 for (int i = 0;i < MAXN; ++ i) if (c[i]){ 31 if (in_degree[i] == 0) re ++; 32 if (in_degree[i] > 1) ok = 0; 33 } 34 if (re == 1 && ok) printf("Case %d is a tree.\n",T); 35 else printf("Case %d is not a tree.\n",T); 36 T++; 37 } 38 return 0; 39 }