Sicily Connect components in undirected graph
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Source:
http://soj.sysu.edu.cn/show_problem.php?pid=1002&cid=2104
Description
输入一个简单无向图,求出图中连通块的数目。
Sample Input
输入的第一行包含两个整数n和m,n是图的顶点数,m是边数。1<=n<=1000,0<=m<=10000。
以下m行,每行是一个数对v y,表示存在边(v,y)。顶点编号从1开始。
5 31 21 32 4
Sample Output
单独一行输出连通块的数目。
2
Caution:
水题,bfs之。
示例代码:
#include<iostream>#include<algorithm>#include<cstring>#include<queue>#include <vector>using namespace std;int n, m, v, y,ans;vector<int> g[10001];int vis[10001];void init(){ cin >> n >> m; while (m--) { cin >> v >> y; g[v].push_back(y); g[y].push_back(v); }}void bfs(int s){ vis[s] = 1; queue<int> q; q.push(s); while (!q.empty()) { int u = q.front(); for (int i = 0; i < g[u].size(); ++i) if (!vis[g[u][i]]) { q.push(g[u][i]); vis[g[u][i]] = 1; } q.pop(); }}void solve(){ for (int i = 1; i <= n; ++i) { if (vis[i] != 1) { ++ans; bfs(i); } } cout << ans << endl;}int main(){ init(); solve(); return 0;}
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