标签：roo with last string put void following collect which
question：
InputThe input will consist of a sequence of descriptions (test cases) followed by a pair of negative integers. Each test case will consist of a sequence of edge descriptions followed by a pair of zeroes Each edge description will consist of a pair of integers; the first integer identifies the node from which the edge begins, and the second integer identifies the node to which the edge is directed. Node numbers will always be greater than zero.
OutputFor each test case display the line ``Case k is a tree." or the line ``Case k is not a tree.", where k corresponds to the test case number (they are sequentially numbered starting with 1).
Sample Input
6 8 5 3 5 2 6 4 5 6 0 0 8 1 7 3 6 2 8 9 7 5 7 4 7 8 7 6 0 0 3 8 6 8 6 4 5 3 5 6 5 2 0 0 -1 -1
Sample Output
Case 1 is a tree. Case 2 is a tree. Case 3 is not a tree.
题意： 考查的是并查集和有向树，套用模板
#include<cstdio> #include<cstring> using namespace std; int parent[1010]; int mm[1010]; int flag; int find_root(int x){ while(parent[x]!=x){ int r=parent[x]; parent[x]=parent[r]; x=r; } return x; } void union_vertices(int x,int y){ int x_root=find_root(x); int y_root=find_root(y); if(x_root!=y_root) parent[y_root]=x_root;//注意 } int main(){ int x,y; int ant=1; while(1){ flag=0; memset(mm,0,sizeof(mm)); for(int i=1;i<=1005;i++) parent[i]=i; while(scanf("%d%d",&x,&y)){ if(x<0&&y<0) return 0; if(x==0&&y==0) break; if(find_root(x)==find_root(y)||find_root(y)!=y) flag=1; else union_vertices(x,y); mm[x]=1,mm[y]=1; } int cnt=0; for(int i=1;i<=1005;i++) if(mm[i]&&find_root(i)==i) cnt++; if(cnt>1) flag=1; if(flag) printf("Case %d is not a tree.\n",ant++); else printf("Case %d is a tree.\n",ant++); } return 0; }
hdu1325 Is It A Tree? solution
标签：roo with last string put void following collect which
原文地址：https://www.cnblogs.com/hrlsm/p/13381815.html
