NYOJ 920 Trees

Trees

时间限制：1000 ms | 内存限制：65535 KB

难度：2

描述

A graph consists of a set 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 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.

Give a graph, report the number of connected components that are also trees.

输入

The input consist of a number of cases. Each case starts with two non-negative integer n and m, satisfying n <= 500 and m <= n(n-1)/2. This is followed by m lines,each containing two integers specifying the two vertices connected by an edge. The vertices are labeled from 1 to n. The end of input is indicated by a line containing n = m = 0.

输出

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).
　　

样例输入

样例输出

Case 1: A forest of 3 trees.
Case 2: There is one tree.
Case 3: No Trees.

题意：给出一张由n个点和m条边构成的无向图，不是连通的，判断图的每一部分是否是一个树，即图中有几棵树。

满足下列条件可以的点和边可以构成一棵树：1.n个点由n-1条边相连，任意两个点之间只有一条边相连。

解题思路：用并查集把各个部分找出来，对于每一部分判断所有点的度之和与点个数的关系，

如果度之和等于点数*2-2，则可以构成一棵树。

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N = 5e2 + 10;
int father[N], deg[N];
vector<int> vec[N];void Initial(int n) //初始化
{for(int i = 1; i <= n; i++)father[i] = i, deg[i] = 0;
}int Find(int x) //寻找父节点
{if(father[x] != x)father[x] = Find(father[x]);return father[x];
}void Union(int a, int b) //合并两个集合
{int p = Find(a);int q = Find(b);if(p != q)father[p] = q;
}int main()
{int n, m, i, j, cas = 1;while(~scanf("%d%d",&n,&m) && (n+m)){Initial(n); //并查集初始化int u, v;for(i = 0; i < m; i++){scanf("%d%d",&u,&v);deg[u]++;deg[v]++;Union(u,v);} //求每个点的度数for(i = 1; i <= n; i++){vec[i].clear();//删除容器中保存的所有元素if(Find(i) == i){for(j = 1; j <= n; j++){if(Find(j) == Find(i))vec[i].push_back(j);}}}//找出哪些点属于同一个集合int ans = 0;for(i = 1; i <= n; i++){int cnt = vec[i].size();if(cnt == 0)continue;int sum = 0;for(j = 0; j < cnt; j++){sum += deg[vec[i][j]];}//求集合中点的度数之和if(sum == cnt * 2 - 2)ans++;  //图中无环}printf("Case %d: ",cas++);if(ans == 0)printf("No Trees.\n");else if(ans == 1)printf("There is one tree.\n");elseprintf("A forest of %d trees.\n",ans);}return 0;
}

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NYOJ 920 Trees

Trees

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