hnsd11348tree(并查集)
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A graph consists of a set of vertices and edges between pairs of vertices. Two vertices are connected if there is a path (subset of edges) leading from one vertex to another, and a connected component is a maximal subset of vertices that are all connected to each other. A graph consists of one or more connected components.
A tree is a connected component without cycles, but it can also be characterized in other ways. For example, a tree consisting of n vertices has exactly n-1 edges. Also, there is a unique path connecting any pair of vertices in a tree.
Given a graph, report the number of connected components that are also trees.
Input
The input consists of a number of cases. Each case starts with two non-negative integersn and m, satisfying n ≤ 500 and m ≤ n(n-1)/2. This is followed by m lines, each containing two integers specifying the two distinct vertices connected by an edge. No edge will be specified twice (or given again in a different order). The vertices are labelled 1 to n. The end of input is indicated by a line containingn = m = 0.
Output
For each case, print one of the following lines depending on how many different connected components are trees (T > 1 below):
Case x: A forest of T trees.Case x: There is one tree.Case x: No trees.
x is the case number (starting from 1).
Sample Input
6 31 22 33 46 51 22 33 44 55 66 61 22 31 34 55 66 40 0Sample Output
Case 1: A forest of 3 trees.Case 2: There is one tree.Case 3: No trees.
#include<stdio.h>int fath[505],cycl[505],k,n;void setfirst(){ k=n; for(int i=1;i<=n;i++) { fath[i]=i; cycl[i]=0; }}int find_fath(int x){ if(x!=fath[x]) fath[x]=find_fath(fath[x]); return fath[x];}void setTree(int a,int b){ a=find_fath(a); b=find_fath(b); if(cycl[b]&&cycl[a]) return ; k--; if(a!=b) { if(cycl[a]) fath[b]=a; else fath[a]=b; } else cycl[a]=1;}int main(){ int a,b,m,t=1; while(scanf("%d%d",&n,&m)>0&&m+n!=0) { setfirst(); while(m--) { scanf("%d%d",&a,&b); setTree(a,b); } printf("Case %d: ",t++); if(k>1)printf("A forest of %d trees.\n",k); if(k==1)printf("There is one tree.\n"); if(k==0)printf("No trees.\n"); }}
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