Trees

時間限制：1000?ms ?|? 內存限制：65535?KB 難度：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.? 輸入

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.

輸出

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).
　　

樣例輸入

6 3 1 2 2 3 3 4 6 5 1 2 2 3 3 4 4 5 5 6 6 6 1 2 2 3 1 3 4 5 5 6 6 4 0 0

樣例輸出

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

題意：給出一張由n個點和m條邊構成的無向圖，不是連通的，判斷圖的每一部分是否是一個樹，即圖中有幾棵樹。

滿足下列條件可以的點和邊可以構成一棵樹：1.n個點由n-1條邊相連，任意兩個點之間只有一條邊相連。

解題思路：用并查集把各個部分找出來，對于每一部分判斷所有點的度之和與點個數的關系，

如果度之和等于點數*2-2，則可以構成一棵樹。

#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) //尋找父節點 {if(father[x] != x)father[x] = Find(father[x]);return father[x]; }void Union(int a, int b) //合并兩個集合 {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);} //求每個點的度數for(i = 1; i <= n; i++){vec[i].clear();//刪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);}}}//找出哪些點屬于同一個集合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]];}//求集合中點的度數之和if(sum == cnt * 2 - 2)ans++; //圖中無環}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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