BFS(60)分

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>

using namespace std;

const int N = 1010,M = N*N;

typedef pair<int,int> PII;

int n,m;
int h[N],e[M],ne[N],idx;
bool g[N][N];
bool st[N];
queue<PII> q;

void add(int a,int b)
{
    e[idx] = b,ne[idx] = h[a],h[a] = idx++;
}

void bfs()
{
    while(q.size())
    {
        auto t = q.front();
        q.pop();

        for(int i = h[t.second]; i!=-1; i = ne[i])
        {
            int j = e[i];
            if(g[t.first][j]) continue;
            q.push({t.first,j});
            g[t.first][j] = true;
        }
    }
}

int main()
{
    cin>>n>>m;
    memset(h, -1, sizeof h);
    for(int i = 0; i<m; i++)
    {
        int a,b;
        cin>>a>>b;
        add(a,b);
        q.push({a,b});
    }

    bfs();

    int res = 0;
    for(int i = 1; i <= n; i ++)
    {
        for(int j = 1; j < i; j ++)
        {
            if(g[i][j] && g[j][i]) res++;

        }
    }
    cout<<res<<endl;
    return 0;
}

DFS(60分)

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
// dfs 60分

const int N = 10000;
int n, m;
int h[N], e[N], ne[N], idx, res;
bool visited[N][N];

void add(int a, int b)
{
    e[idx] = b;
    ne[idx] = h[a];
    h[a] = idx++;
}

void dfs(int root, int cur)
{
    visited[root][cur] = true;
    for (int i = h[cur]; ~i; i = ne[i])
    {
        int neighbor = e[i];
        if (visited[root][neighbor])
            continue;
        dfs(root, neighbor);
    }
}

int main()
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    while (m--)
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
    }

    for (int i = 1; i <= n; i++)
    {
        dfs(i, i);
    }

    res = 0;

    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j < i; j++)
        {
            if (visited[i][j] && visited[j][i])
                res++;
        }
    }
    cout << res << endl;
    return 0;
}

有向图强连通分量(板子),提高课1174

#include <iostream>
#include <cstring>

using namespace std;

typedef long long LL;

const int N = 1e4+10, M = 1e5+10;

int n,m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N];
int stk[N], top,ts;
bool in_stk[N];
int ans;

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

//有向图强连通分量，维护每个连通块内点的数量
void tarjan(int u)
{
    dfn[u] = low[u] = ++ts;
    stk[++top] = u;
    in_stk[u] = true;

    for(int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(!dfn[j])
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if(in_stk[j]) low[u] = min(low[u], dfn[j]);
    }

    if(dfn[u] == low[u])
    {
        int y,cnt = 0;
        do{
            y = stk[top--];
            in_stk[y] = false;
            cnt++;
        }while(y != u);
        ans += cnt*(cnt-1)/2;
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    memset(h, -1, sizeof h);

    while(m--)
    {
        int a,b;
        scanf("%d%d", &a, &b);
        add(a,b);
    }

    for(int i = 1; i <= n; i++)
        if(!dfn[i])
            tarjan(i);
    cout<<ans;
    return 0;
}