求连通块的个数代码 dfs/并查集

dfs代码:

//求连通块的个数
#include<iostream>
#include<cstdio>
#include<vector>
#define maxn 1000
using namespace std;

vector<int> G[maxn];
bool vis[maxn] = {false};

int m, n;

void dfs(int v)
{
    vis[v] = true;
    for(int i = 0; i < G[v].size(); i++)
    {
        if(vis[G[v][i]] == false)
        {
            dfs(G[v][i]);
        }
    }
}


int main()
{
    scanf("%d %d",&m, &n);
    for(int i = 0; i < n; i++)
    {
        int a, b;
        scanf("%d %d", &a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
    }

    int block = 0;
    for(int i = 1; i <= m; i++)
    {
        if(vis[i] == false)
        {
            dfs(i);
            block++;
        }
    }

    printf("%d", block);

    return 0;
}

并查集代码:

#include<iostream>
#include<vector>
#define maxn 1000
using namespace std;

vector<int> G[maxn];
int father[maxn];
bool isroot[maxn];

int findfather(int x)
{
    int a = x;
    while(x != father[x])
    {
        x = father[x];
    }

    //路径压缩
    while(a != father[a])
    {
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}

void Union(int x, int y)
{
    int fx = findfather(x);
    int fy = findfather(y);
    if(fx != fy)
    {
        father[fx] = fy;
    }
}

void init()
{
    for(int i = 1; i < maxn; i++)
    {
        father[i] = i;
        isroot[i] = false;
    }
}

int m ,n;
int main()
{
    scanf("%d %d",&m, &n);
    for(int i = 0; i < n; i++)
    {
        int a, b;
        scanf("%d %d",&a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
    }

    init();
    for(int i = 1; i <= m; i ++)
    {
        for(int j = 0; j < G[i].size(); j++)
        {
            int u = i, v = G[i][j];
            Union(u, v);
        }
    }

    int countnum = 0;
    for(int i = 1; i <= m; i++)
    {
        if(isroot[findfather(i)] == false)
        {
            isroot[findfather(i)] = true;
            countnum++;
        }
    }

    printf("%d",countnum);
    return 0;
}

输入样例:

8 5
1 2
1 3
4 5
4 6
7 8

输出结果:

3

具体解释,有时间再补,先贴上代码备忘

猜你喜欢

转载自blog.csdn.net/young_Tao/article/details/81782031