首页 algorithm 杂杂的算法知识

# 缩点

P3387 【模板】缩点 - 洛谷

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#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e4 + 10, M = 1e5 + 10;
int head1[M], ne1[M], to1[M], from1[M], idx1;
int head2[M], ne2[M], to2[M], from2[M], idx2;
int a[N];
int dfn[N], low[N], sta[N], vis[N], cnt, top;
int f[N];//记录每个节点所在的强连通分量
int in[N], dist[N];
int n, m;
void add(int x, int y)
{
    ne1[idx1] = head1[x], to1[idx1] = y, from1[idx1] = x, head1[x] = idx1++;
}
void add2(int x, int y)
{
    ne2[idx2] = head2[x], to2[idx2] = y, from2[idx2] = x, head2[x] = idx2++;
}
void Tarjan(int u)
{
    dfn[u] = low[u] = ++cnt;
    sta[++top] = u, vis[u] = 1;
    for(int i = head1[u]; ~i; i = ne1[i])
    {
        int v = to1[i];
        if(!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        if(vis[v])//如果在队里
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    //强连通分量的代表元素
    if(dfn[u] == low[u])
    {
        int y;
        while(y = sta[top--])
        {
            f[y] = u;
            vis[y] = 0;
            if(u == y) break;
            a[u] += a[y];
        }
    }
}
//拓扑排序求最长路径
int topo()
{
    queue<int> q;
    for(int i = 1; i <= n; i++)
    {
        if(f[i] == i && in[i] == 0)
        {
            q.push(i);
            dist[i] += a[i];
        }
    }
    while(q.size())
    {
        int t = q.front();
        q.pop();
        for(int i = head2[t]; ~i; i = ne2[i])
        {
            int v = to2[i];
            dist[v] = max(dist[v], dist[t] + a[v]);
            in[v]--;
            if(in[v] == 0) q.push(v);
        }
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) ans = max(ans, dist[i]);
    return ans;
}
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    memset(head1, -1, sizeof head1);
    memset(head2, -1, sizeof head2);
    cin >> n >> m;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = 1; i <= m; i++) 
    {
        int u, v;
        cin >> u >> v;
        add(u, v);
    }
    //Tarjan
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) Tarjan(i);
    //建立缩点之后的图
    for(int i = 1; i <= m; i++)
    {
        int x = f[from1[i]], y = f[to1[i]];
        if(x == y) continue;
        add2(x, y);
        in[y]++;
    }
    cout << topo();
    return 0;
}

# 割点

P3388 【模板】割点（割顶） - 洛谷

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#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e4 + 10, M = 1e5 + 10;
vector<int> e[N];
int dfn[N], low[N], idx;
int buc[N];
int root, cnt;
int n, m;

void tarjan(int u)
{
    dfn[u] = low[u] = ++idx;
    int son = 0;
    for(auto v: e[u])
    {
        if(!dfn[v])
        {
            son++;
            tarjan(v);
            low[u] = min(low[v], low[u]);
            //判断是否是割点
            //1. 不是根节点
            //2. 子节点的low值不小于当前节点的dfn值
            if(root != u && low[v] >= dfn[u])
            {
                cnt += !buc[u], buc[u] = 1;
            }
        }
        else
        {
            low[u] = min(low[u], dfn[v]);
        }
    }
    if(son >= 2 && u == root)
        cnt += !buc[u], buc[u] = 1;
}
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    cin >> n >> m;
    for(int i = 1; i <= m; i++)
    {
        int x, y;
        cin >> x >> y;
        e[x].push_back(y);
        e[y].push_back(x);
    }
    for(int i = 1; i <= n; i++)
        if(!dfn[i]) root = i, tarjan(i);
    cout << cnt << '\n';
    for(int i = 1; i <= n; i++)
        if(buc[i]) cout << i << " ";
    return 0;
}