寒假:Day25
作者:互联网
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;
}
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;
}
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;
}
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;
}
标签:dist,idx,fa,Day25,++,int,depth,寒假 来源: https://blog.csdn.net/hjh2020012124/article/details/122772873