题目

BFS暴力代码 60p

#include <bits/stdc++.h>
using namespace std;

const int N = 1010;
const int M = 10010; 

int g[N][N];
int n, m;
int h[N], e[M], ne[M], idx;

void add(int a, int b)  // 添加一条边a->b
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void bfs(int u)
{
    queue<int> q;
    q.push(u);
    
    while(q.size())
    {
        int t = q.front(); q.pop();
        for(int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if(g[u][j]) continue;
            g[u][j] = 1;
            q.push(j);
        }
    }
}
int main()
{
    memset(h, -1, sizeof h);
    
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    
    for(int i = 1; i <= n; i++)
        bfs(i);
        
    int ans = 0;
    for(int i = 1; i <= n; i++)
        for(int j = i+1; j <= n; j++)
            if(g[i][j] && g[j][i]) ans++;
    cout << ans;
}

Tarjan代码

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

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

int h[N], e[M], ne[M], idx; //链式前向星
int dfn[N], low[N], tot; //时间戳 追溯值 分配器
int scc[N], sz[N], cnt; //强连通分量归属 大小 分配器
int stk[N], tt; //栈相关
bool in[N];
int n, m;

void add(int a, int b)  // 添加一条边a->b
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void tarjan(int u)
{
    dfn[u] = ++tot, low[u] = tot;
    stk[++tt] = u; in[u] = 1;
    
    for(int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if(!dfn[j]) //未访问过，先递归，再更新追溯值
        {
            tarjan(j);
            low[u] = min(low[u], low[j]);
        }
        else if(in[j]) //儿子访问过，但仍在栈中，说明成环，拿时间戳
        {
            low[u] = min(low[u], dfn[j]);
        }
    }
    
    if(dfn[u] == low[u]) //时间戳与追溯值一致，x为强连通分量的根节点，开始退栈
    {
        int t; ++cnt;
        do
        {
            t = stk[tt--]; in[t] = 0;
            scc[t] = cnt;
            sz[cnt]++;
        }while(t != u);
    }
}
int main()
{
    memset(h, -1, sizeof h);
    
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= m; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) tarjan(i);
    
    ll ans = 0;
    for(int i = 1; i <= cnt; i++)
        ans += 1ll * sz[i] * (sz[i] - 1) / 2;
        
    cout << ans;
}