题解 | 畅通工程

// 求连通分量的个数
#include <string.h>
#include <iostream>
#include <set>

using std::cout;
using std::cin;
using std::endl;
using std::set;

bool visited[1000] = {false};
set<int> graph[1000];

void DFS(int i) {
    visited[i] = true;
    for (auto elem : graph[i]) {
        if (!visited[elem])
            DFS(elem);
    }
}

void ClearGraph(int n) {
    for (int i = 1; i <= n; ++i) {
        graph[i].clear();
    }
}

int main() {
    int n, m;
    int t1, t2; // 两个镇子之间有一条路
    while (cin >> n >> m) {
        for (int i = 1; i <= m; ++i) {
            scanf("%d%d", &t1, &t2);
            graph[t1].insert(t2);
            graph[t2].insert(t1);
        }
        int count = 0;
        for (int i = 1; i <= n; ++i) {
            if (!visited[i]) {
                // DFS次数就是连通分量个数
                DFS(i);
                count++;
            }
        }
        cout << count - 1 << endl;	// 连通分量个数-1就是需要再铺建的道路的数量
        memset(visited, false, sizeof(visited));
        ClearGraph(n);
    }

    return 0;
}