Day25

最近公共祖先问题(LCA)1172. 祖孙询问 - AcWing题库

LCA模板题

#include<bits/stdc++.h>
using namespace std;
const int N = 40010, M = 2 * N;

int n, m;
int h[N], e[M], ne[M], idx;
int depth[N], fa[N][16]; // depth存每个节点的深度，fa存倍增往前走2^i步到哪个点编号
int q[N]; // 数组模拟队列

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

void bfs(int root) // 宽搜遍历图预处理fa数组，顺便预处理depth数组
{
    memset(depth, 0x3f, sizeof depth); 
    depth[0] = 0, depth[root] = 1; // 根节点深度为1，越界的话为0
    int hh = 0, tt = 0;
    q[0] = root;
    while (hh <= tt)
    {
        int t = q[hh++];
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (depth[j] > depth[t] + 1)
            {
                depth[j] = depth[t] + 1;
                q[++tt] = j;
                fa[j][0] = t; // j的父节点是t
                for (int k = 1; k <= 15; k++)
                    fa[j][k] = fa[fa[j][k - 1]][k - 1]; // 递推求fa数组
            }
        }
    }
}

int lca(int a, int b)
{
    if (depth[a] < depth[b]) swap(a, b); // 保证a的深度大于b
    
    for (int k = 15; k >= 0; k--)
        if (depth[fa[a][k]] >= depth[b]) // a往上跳到和b同一个深度
            a = fa[a][k];
            
    if (a == b) return a; // 如果a和b是一个节点，直接返回即可
    for (int k = 15; k >= 0; k--)
        if (fa[a][k] != fa[b][k]) // a和b同时往上跳，直到跳到公共祖先的子节点
        {
            a = fa[a][k];
            b = fa[b][k];
        }
    return fa[a][0]; // 返回父节点即可
}

int main(void)
{
    cin >> n;
    int root;
    memset(h, -1, sizeof h);
    
    for (int i = 0; i < n; i++)
    {
        int a, b;
        cin >> a >> b;
        if (b == -1) root = a;
        else add(a, b), add(b, a);
    }
    
    bfs(root);
    
    cin >> m;
    while (m--)
    {
        int a, b;
        cin >> a >> b;
        int p = lca(a, b);
        if (p == a) puts("1");
        else if (p == b) puts("2");
        else puts("0");
    }
    return 0;
}

1171. 距离 - AcWing题库

Tarjan离线算法求LCA，将所有询问读入后一次性输出

#include<bits/stdc++.h>
using namespace std;
const int N = 20010, M = N * 2;

typedef pair<int, int> PII;

int n, m;
int h[N], e[M], w[M], ne[M], idx;
int dist[N]; // 当前节点到根节点的距离
int p[N];
int res[N];
int st[N];
vector<PII> query[N]; // first存查询的另外一个点， second存查询编号

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

void dfs(int u, int fa) // 深搜预处理dist数组
{
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j == fa) continue;
        dist[j] = dist[u] + w[i];
        dfs(j, u);
    }
}

int find(int x)
{
    if (x != p[x]) p[x] = find(p[x]);
    return p[x];
}

void tarjan(int u)
{
    st[u] = 1; 
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (!st[j]) // 如果当前节点没有被遍历
        {
            tarjan(j);
            p[j] = u; // 把当前点的祖宗节点更新为u
        }
    }
    for (auto item : query[u]) // 如果搜到叶子节点了，就开始更新答案
    {
        int y = item.first, id = item.second;
        if (st[y] == 2) // 
        {
            int anc = find(y); // 找到y的祖宗结点
            res[id] = dist[u] + dist[y] - dist[anc] * 2; // 更新距离
        }
    }
    
    st[u] = 2; // 把当前节点置为2，表示已经搜过
}

int main(void)
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    for (int i = 0; i < n - 1; i++)
    {
        int a, b, c;
        cin >> a >> b >> c;
        add(a, b, c), add(b, a, c);
    }
    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        if (a != b)
        {
            query[a].push_back({b, i});
            query[b].push_back({a, i});
        }
    }
    for (int i = 1; i <= n; i++) p[i] = i;
    dfs(1, -1); // 任取一个根节点，然后把fa置为-1即可
    tarjan(1);
    for (int i = 0; i < m; i++) cout << res[i] << endl;
    return 0;
}

356. 次小生成树 - AcWing题库

利用LCA求次小生成树：

先建出最小生成树，然后枚举非树边，利用LCA求出非树边两端之间的最大边和次大边，最后返回的是和非树边的差值，最后求 m i n min min 即可

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 100010, M = 300010, INF = 0x3f3f3f3f;

int n, m;
struct Edge
{
    int a, b, w;
    bool used;
    bool operator< (const Edge &t) const // 重载小于号排序
    {
        return w < t.w;
    }
}edge[M];
int p[N]; // 并查集数组
int h[N], e[M], w[M], ne[M], idx; // 邻接表
int depth[N], fa[N][17], d1[N][17], d2[N][17]; // d1最大边，d2次大边
int q[N]; // 队列

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

int find(int x)
{
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

LL kruskal() // 先求出最小生成树，并把树边标记上
{
    for (int i = 1; i <= n; i ++ ) p[i] = i;
    sort(edge, edge + m);
    LL res = 0;
    for (int i = 0; i < m; i ++ )
    {
        int a = find(edge[i].a), b = find(edge[i].b), w = edge[i].w;
        if (a != b)
        {
            p[a] = b;
            res += w;
            edge[i].used = true;
        }
    }
    return res;
}

void build() // 把最小生成树建出来
{
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i ++ )
        if (edge[i].used)
        {
            int a = edge[i].a, b = edge[i].b, w = edge[i].w;
            add(a, b, w), add(b, a, w);
        }
}

void bfs() // 宽搜预处理depth数组和fa数组
{
    memset(depth, 0x3f, sizeof depth);
    depth[0] = 0, depth[1] = 1; // 0是哨兵，1是根节点
    q[0] = 1;
    int hh = 0, tt = 0;
    while (hh <= tt)
    {
        int t = q[hh ++ ];
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (depth[j] > depth[t] + 1)
            {
                depth[j] = depth[t] + 1;
                q[ ++ tt] = j;
                fa[j][0] = t;
                d1[j][0] = w[i], d2[j][0] = -INF;
                for (int k = 1; k <= 16; k ++ )
                {
                    int anc = fa[j][k - 1];
                    fa[j][k] = fa[anc][k - 1];
                    int dist[4] = {d1[j][k - 1], d2[j][k - 1], d1[anc][k - 1], d2[anc][k - 1]};
                    d1[j][k] = d2[j][k] = -INF;
                    for (int u = 0; u < 4; u ++ )
                    {
                        int d = dist[u];
                        if (d > d1[j][k]) d2[j][k] = d1[j][k], d1[j][k] = d;
                        else if (d != d1[j][k] && d > d2[j][k]) d2[j][k] = d;
                    }
                }
            }
        }
    }
}

int lca(int a, int b, int w)
{
    static int dist[N * 2];
    int cnt = 0;
    if (depth[a] < depth[b]) swap(a, b);
    for (int k = 16; k >= 0; k -- )
        if (depth[fa[a][k]] >= depth[b])
        {
            dist[cnt ++ ] = d1[a][k];
            dist[cnt ++ ] = d2[a][k];
            a = fa[a][k];
        }
    if (a != b)
    {
        for (int k = 16; k >= 0; k -- )
            if (fa[a][k] != fa[b][k])
            {
                dist[cnt ++ ] = d1[a][k];
                dist[cnt ++ ] = d2[a][k];
                dist[cnt ++ ] = d1[b][k];
                dist[cnt ++ ] = d2[b][k];
                a = fa[a][k], b = fa[b][k];
            }
        dist[cnt ++ ] = d1[a][0];
        dist[cnt ++ ] = d1[b][0];
    }

    int dist1 = -INF, dist2 = -INF;
    for (int i = 0; i < cnt; i ++ )
    {
        int d = dist[i];
        if (d > dist1) dist2 = dist1, dist1 = d;
        else if (d != dist1 && d > dist2) dist2 = d;
    }

    if (w > dist1) return w - dist1;
    if (w > dist2) return w - dist2;
    return INF;
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < m; i ++ )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        edge[i] = {a, b, c};
    }
    
    LL sum = kruskal();
    build();
    bfs();

    LL res = 1e18;
    for (int i = 0; i < m; i ++ )
        if (!edge[i].used)
        {
            int a = edge[i].a, b = edge[i].b, w = edge[i].w;
            res = min(res, sum + lca(a, b, w));
        }
    cout << res << endl;
    return 0;
}

352. 闇の連鎖 - AcWing题库

树上差分：

在一颗树上加边，必然会形成环，我们就对这条边两个端点求最近公共祖先，然后 d [ a ] + 1 , d [ b ] + 1 , d [ p ] − 2 d[a]+1, d[b]+1, d[p] - 2 d[a]+1,d[b]+1,d[p]−2 然后深搜求解即可

#include<bits/stdc++.h>
using namespace std;
const int N = 100010, M = 2 * N;

int n, m;
int h[N], e[M], ne[M], idx;
int depth[N], fa[N][17];
int d[N];
int q[N];
int ans;

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

void bfs() // 预处理depth数组和fa数组
{
    memset(depth, 0x3f, sizeof depth);
    depth[0] = 0, depth[1] = 1;
    int hh = 0, tt = 0;
    q[0] = 1;
    while (hh <= tt)
    {
        int t = q[hh++];
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (depth[j] > depth[t] + 1)
            {
                depth[j] = depth[t] + 1;
                q[++tt] = j;
                fa[j][0] = t;
                for (int k = 1; k <= 16; k++)
                    fa[j][k] = fa[fa[j][k - 1]][k - 1];
            }
        }
    }
}

int lca(int a, int b) // lca求最近公共祖先
{
    if (depth[a] < depth[b]) swap(a, b);
    for (int k = 16; k >= 0; k--)
        if (depth[fa[a][k]] >= depth[b])
            a = fa[a][k];
    if (a == b) return a;
    for (int k = 16; k >= 0; k--)
        if (fa[a][k] != fa[b][k])
        {
            a = fa[a][k];
            b = fa[b][k];
        }
    return fa[a][0];
}

int dfs(int u, int fa) // 树上差分
{
    int res = d[u];
    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j != fa)
        {
            int s = dfs(j, u);
            if (s == 0) ans += m;
            else if (s == 1) ans++;
            res += s;
        }
    }
    return res;
}

int main(void)
{
    cin >> n >> m;
    memset(h, -1, sizeof h);
    for (int i = 0; i < n - 1; i++)
    {
        int a, b;
        cin >> a >> b;
        add(a, b), add(b, a);
    }
    bfs();
    for (int i = 0; i < m; i++)
    {
        int a, b;
        cin >> a >> b;
        int p = lca(a, b);
        d[a]++, d[b]++, d[p] -= 2;
    }
    dfs(1, -1);
    cout << ans << endl ;
    return 0;
}