标签:pre dist int rep cin DFS 序和树 void 欧拉
倍增
定义
倍增法,顾名思义就是翻倍.
它能够使线性的处理转化为对数级的处理,大大地优化时间复杂度
这个方法在很多算法中均有应用,其中最常用的是 RMQ 问题和求LCA,无修改的路径信息。
路径最小值
注意:路径上的信息需要可以合并,例如求最值
const int N = 201000;
const int LOGN = 18;
int n, q;
int dep[N], par[N][LOGN + 1], val[N][LOGN + 1];
vector< pair<int,int> > e[N];
il void dfs(int u, int f)
{
dep[u] = dep[f] + 1;
for (auto p : e[u])
{
int v = p.fi;
if (v == f) continue;
par[v][0] = u;
val[v][0] = p.second;
dfs(v, u);
}
}
int query(int u, int v)
{
int ans = 1 << 30;
if (dep[u] > dep[v]) swap(u, v);
int d = dep[v] - dep[u];
per(j,LOGN,0) if (d & (1 << j))
{
ans = min(ans, val[v][j]);
v = par[v][j];
}
if (u == v) return ans;
per(j,LOGN,0) if (par[u][j] != par[v][j])
{
ans = min(ans, min(val[u][j], val[v][j]));
u = par[u][j];
v = par[v][j];
}
ans = min({ans, val[u][0], val[v][0]});
return ans;
}
int main(void)
{
clock_t c1 = clock();
FastIO();
cin >> n >> q;
rep(i,1,n-1)
{
int u, v, w;
cin >> u >> v >> w;
e[u].pb(make_pair(v,w));
e[v].pb(make_pair(u,w));
}
dfs(1,0);
//预处理掉这部分数据
rep(j,1,LOGN) rep(u,1,n)
{
par[u][j] = par[par[u][j-1]][j-1];
val[u][j] = min(val[u][j-1], val[par[u][j-1]][j-1]);
}
rep(i,1,q)
{
int u, v;
cin >> u >> v;
cout << query(u, v) << '\n';
}
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
DFS序
定义
树的DFS序列,也就是树的深搜序,它的概念是:树的每一个节点在深度优先遍历中进出栈的时间序列。
作用
可以把子树问题转化为序列问题,非线性问题变成线性问题求解,可以用上树状数组或者线段树等数据结构进行信息维护。
可以用来做带修改的子树信息这一类题目。
例题1
题面
给一颗n个点的树,有两个操作。
- 1 x y, 把x的点权修改成y
- 2 x,询问x点的子树点权和和到根的路径的点权和(含x点)
代码实现
两个树状数组,分别维护前缀和和差分。
前缀和数组用来维护子树点权和,差分数组用来维护到根的路径点全和。
const int N = 2e5 + 10;
template<class T>
struct BIT
{
T c[N];
int size;
void resize(int s)
{
size = s;
rep(i,1,size) c[i] = 0;
}
void modify(int x, T d)
{
assert(x != 0);
for (; x <= size; x += x & (-x)) c[x] += d;
}
T query(int x)
{
assert(x<=size);
T s = 0;
for (; x; x -= x & (-x)) s += c[x];
return s;
}
};
int n, q, tot;
int a[N],l[N],r[N];
vector<int> e[N];
BIT<ll> c1, c2;
il void dfs(int u, int f)
{
l[u] = ++tot;
for (auto v : e[u]) if (v == f) continue; else dfs(v, u);
r[u] = tot;
}
int main(void)
{
// clock_t c1 = clock();
FastIO();
cin >> n >> q;
rep(i,2,n)
{
int u, v;
cin >> u >> v;
e[u].pb(v), e[v].pb(u);
}
dfs(1, 0);
c1.resize(n), c2.resize(n);
rep(i,1,n)
{
cin >> a[i];
c1.modify(l[i], a[i]);
c2.modify(l[i], a[i]);
c2.modify(r[i] + 1, -a[i]);
}
rep(i,1,q)
{
int ty; cin >> ty;
if (ty == 1)
{
int x, y;
cin >> x >> y;
int d = y - a[x];
a[x] = y;
c1.modify(l[x], d);
c2.modify(l[x], d);
c2.modify(r[x] + 1, -d);
}
else
{
int x;
cin >> x;
cout << c1.query(r[x]) - c1.query(l[x] - 1) << " ";
cout << c2.query(l[x]) << '\n';
}
}
// cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
DFS序列2
代码实现
const int N = 2e5 + 10;
template<class T>
struct BIT
{
T c[N];
int size;
void resize(int s)
{
size = s;
rep(i,1,size) c[i] = 0;
}
void modify(int x, T d)
{
assert(x != 0);
for (; x <= size; x += x & (-x)) c[x] += d;
}
T query(int x)
{
assert(x<=size);
T s = 0;
for (; x; x -= x & (-x)) s += c[x];
return s;
}
};
int n, q, tot;
int a[N],l[N],r[N];
vector<int> e[N];
vector< pair<int, int> > son[N];
BIT<ll> c1;
void dfs(int u, int f)
{
l[u] = ++tot;
for (auto v : e[u])
{
if (v == f) continue;
else dfs(v, u);
son[u].pb({l[v], r[v]});
}
r[u] = tot;
}
int main(void)
{
// clock_t c1 = clock();
ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
cin >> n >> q;
rep(i,2,n)
{
int u, v;
cin >> u >> v;
e[u].pb(v), e[v].pb(u);
}
int root = 1;
dfs(1, 0);
c1.resize(n);
rep(i,1,n)
{
cin >> a[i];
c1.modify(l[i], a[i]);
}
rep(i,1,q)
{
int ty; cin >> ty;
if (ty == 1)
{
int x, y;
cin >> x >> y;
int d = y - a[x];
a[x] = y;
c1.modify(l[x], d);
}
else if (ty == 3)
{
cin >> root;
}
else
{
int x;
cin >> x;
if (x == root) cout << c1.query(n) << '\n';
else if (l[x] < l[root] && r[root] <= r[x])
{
auto seg = *prev(upper_bound(all(son[x]), pair<int,int>{l[root], r[root]}));
cout << c1.query(n) - (c1.query(seg.second) - c1.query(seg.first - 1)) << '\n';
}
else cout << c1.query(r[x]) - c1.query(l[x] - 1) << '\n';
}
}
return 0;
}
欧拉序
相比DFS序,多记录一层回溯的点。
const int N = 501000, LOGN = 20;
int n, q, p[N], tot, dep[N];
vector<int> e[N];
PII f[LOGN + 2][N * 2];
ll ans;
unsigned int A, B, C;
inline unsigned int rng61() {
A ^= A << 16;
A ^= A >> 5;
A ^= A << 1;
unsigned int t = A;
A = B;
B = C;
C ^= t ^ A;
return C;
}
void dfs(int u, int par)
{
p[u] = ++tot;
dep[u] = dep[par] + 1;
f[0][tot] = {dep[u], u};
for (auto v : e[u])
{
if (v == par) continue;
dfs(v, u);
f[0][++tot] = {dep[u], u};
}
}
int main(){
scanf("%d%d%u%u%u", &n, &q, &A, &B, &C);
for (int i = 1; i < n; i++)
{
int u, v;
scanf("%d%d",&u,&v);
e[u].pb(v);
e[v].pb(u);
}
dfs(1, 0);
for (int j = 1; j <= LOGN; j++)
for (int i = 1; i + (1 << j) - 1 <= tot; i++)
f[j][i] = min(f[j-1][i], f[j - 1][i + (1 << (j - 1))]);
for (int i = 1; i <= q; i++)
{
int l = p[rng61() % n + 1], r = p[rng61() % n + 1];
if (l > r) swap(l, r);
int len = 31 - __builtin_clz(r - l + 1);
int d = min(f[len][l], f[len][r - (1 << len) + 1]).second;
ans ^= 1ll * i * d;
}
printf("%lld\n",ans);
return 0;
}
树
树的直径
树的直径是指树上任意两个节点之间最长.
树的直径的中间节点被称为树的中心,如果直径上有偶数个点,那么中间的两个节点都可以是树的中心.
树的中心到其他点的最长路径最短.
代码实现
int n, pre[N], c[N], q[N], dist[N];
int l, front = 1, rear = 0;
vector<int> e[N];
il void dfs(int x)
{
for (auto y : e[x])
{
if (y != pre[x])
{
pre[y] = x;
dist[y] = dist[x] + 1;
dfs(y);
}
}
}
int main(void)
{
FastIO();
cin >> n;
rep(i,1,n-1)
{
int u, v;
cin >> u >> v;
e[u].pb(v), e[v].pb(u);
}
memset(dist, 0, sizeof(dist));
memset(pre, 0, sizeof(pre));
pre[1] = -1;
dfs(1);
int idx = 0, v = 0;
rep(i,1,n)
{
if (dist[i] > v) v = dist[i], idx = i;
}
memset(dist, 0, sizeof(dist));
memset(pre, 0, sizeof(pre));
pre[idx] = -1;
dfs(idx);
v = 0;
rep(i,1,n) v = max(v, dist[i]);
cout << v << '\n';
return 0;
}
树的重心
对于一颗无根树而言,当一个节点被选为根节点,它底下的每一个子节点的子树的大小最大值最小的那个点,被称为树的重心
删除重心后,树分裂成若干个子树,这若干个子树中的最大值最小.
性质
- 当重心为根节点时,它底下的每一个子树大小不大于整棵树大小的一半
- 重心到其他所有节点的距离和最小
代码实现
小数据\(O(logn)\)
const int N = 1e5 + 10;
int n, pre[1001], c[N], q[N], dist[N], f[N];
int l, front = 1, rear = 0;
vector<int> e[N];
il void dfs(int x)
{
for (auto y : e[x])
{
if (y != pre[x])
{
pre[y] = x;
dist[y] = dist[x] + 1;
dfs(y);
}
}
}
il void solve(int x)
{
++cnt;
for (auto y : e[x])
if (y != pre[x])
pre[y] = x, solve(y);
}
int main(void)
{
FastIO();
cin >> n;
rep(i,1,n-1)
{
int u, v;
cin >> u >> v;
e[u].pb(v), e[v].pb(u);
}
rep(i,1,n)
{
f[i] = 0;
memset(pre, 0, sizeof(pre));
for (auto y : e[i])
{
cnt = 0;
pre[y] = i;
solve(y);
f[i] = max(f[i], cnt);
}
}
// idx 重心
int idx = 0, v = 1 << 30;
rep(i,1,n) if (f[i] < v) v = f[i], idx = i;
memset(dist, 0, sizeof(dist));
memset(pre, 0, sizeof(pre));
pre[idx] = -1;
dfs(idx);
int ans = 0;
rep(i,1,n) ans += dist[i];
cout << ans << '\n';
return 0;
}
大数据
时间复杂度\(O(logn)\)
#include <bits/stdc++.h>
using namespace std;
#define DABIAO freopen("in.in","r",stdin);freopen("out.out","w",stdout);
#define debug(x...) do { cout << "\033[32;1m" << #x << " --> "; rd_debug(x); } while (0)
void rd_debug() {cout << "\033[39;0m" << endl; }
template<class T, class... Ts> void rd_debug(const T& arg, const Ts&... args) { cout << arg << " "; rd_debug(args...); }
#define FastIO() ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define rep(i, l, r) for(int i = (l); i <= (r); i++)
#define per(i, r, l) for(int i = (r); i >= (l); i--)
#define pb push_back
#define SZ(x) ((int)(x).size())
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define il inline
typedef long long ll;
typedef double db;
typedef pair<int, int> PII;
const int inf = 0x3f3f3f3f;
const ll infl = 0x3f3f3f3f3f3f3f3fll;
const double eps = 1e-6;
const int N = 1e5 + 10;
int n, pre[1001], c[N], q[N], dist[N], f[N];
int l, front = 1, rear = 0;
vector<int> e[N];
il void dfs(int x)
{
for (auto y : e[x])
{
if (y != pre[x])
{
pre[y] = x;
dist[y] = dist[x] + 1;
dfs(y);
}
}
}
il void solve(int x)
{
s[x] = 1;
for (auto y : e[x])
if (y != pre[x])
pre[y] = x, solve(y), s[x] += s[y];
}
int main(void)
{
FastIO();
cin >> n;
rep(i,1,n-1)
{
int u, v;
cin >> u >> v;
e[u].pb(v), e[v].pb(u);
}
memset(pre, 0, sizeof(pre));
pre[1] = -1;
solve(1);
int idx = 0, v = 1 << 30;
rep(i,1,n)
{
int f = 0;
for (auto y : e[i])
{
if (y != pre[i])
f = max(f, s[y]);
else
f = max(f, n - s[i]);
if (f < v) v = f, idx = i;
}
}
memset(dist, 0, sizeof(dist));
memset(pre, 0, sizeof(pre));
pre[idx] = -1;
dfs(idx);
int ans = 0;
rep(i,1,n) ans += dist[i];
cout << ans << '\n';
return 0;
}
LCA
求法
- 先计算两个节点深度
- 调整同一深度
- 两个节点一起往上跳,直到两个节点相等
代码实现
小数据
时间复杂度\(O(n^2)\)
const int N = 1001;
int n, m, father[N], dist[N];
vector<int> e[N];
il void dfs(int x){ for (auto y : e[x]) dist[y] = dist[x] + 1, dfs(y); }
int main(void)
{
clock_t c1 = clock();
FastIO();
cin >> n;
rep(i,1,n)
{
int x, y;
cin >> x >> y;
e[x].pb(y);
father[y] = x;
}
memset(dist, 0, sizeof(dist));
dfs(1);
cin >> m;
rep(i,1,m)
{
int x, y;
cin >> x >> y;
if (dist[x] < dist[y]) swap(x, y);
int z = dist[x] - dist[y];
rep(i,j,z) x = father[x];
while (x != y) x = father[x], y = father[y];
cout << x << '\n';
}
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
大数据
时间复杂度\(O(logn)\)
int n, m, father[N], dist[N];
vector<int> e[N];
il void dfs(int x)
{
for (auto y : e[x]) dist[y] = dist[x] + 1, dfs(y);
}
int main(void)
{
clock_t c1 = clock();
FastIO();
cin >> n;
rep(i,1,n)
{
int x, y;
cin >> x >> y;
e[x].pb(y);
father[y][0] = x;
}
rep(i,1,20) rep(j,1,n)
if (father[j][i - 1])
father[j][i] = father[father[j][i-1]][i-1];
memset(dist, 0, sizeof(dist));
dfs(1);
cin >> m;
rep(i,1,m)
{
int x, y;
cin >> x >> y;
if (dist[x] < dist[y]) swap(x, y);
int z = dist[x] - dist[y];
for (int j = 0; j <= 20 && z; j++, z >>= 1)
if (z & 1) x = father[x][j];
if (x == y) { cout << x << '\n'; continue; }
per(j,20,0) if (father[x][j] != father[y][j])
x = father[x][j], y = father[y][j];
cout << father[x][0] << '\n';
}
cerr << "Time Used:" << clock() - c1 << "ms" << endl;
return 0;
}
用欧拉序来求LCA
const int N = 501000, LOGN = 20;
int n, q, p[N], tot, dep[N];
vector<int> e[N];
PII f[LOGN + 2][N * 2];
ll ans;
unsigned int A, B, C;
inline unsigned int rng61() {
A ^= A << 16;
A ^= A >> 5;
A ^= A << 1;
unsigned int t = A;
A = B;
B = C;
C ^= t ^ A;
return C;
}
void dfs(int u, int par)
{
p[u] = ++tot;
dep[u] = dep[par] + 1;
f[0][tot] = {dep[u], u};
for (auto v : e[u])
{
if (v == par) continue;
dfs(v, u);
f[0][++tot] = {dep[u], u};
}
}
int main(){
scanf("%d%d%u%u%u", &n, &q, &A, &B, &C);
for (int i = 1; i < n; i++)
{
int u, v;
scanf("%d%d",&u,&v);
e[u].pb(v);
e[v].pb(u);
}
dfs(1, 0);
for (int j = 1; j <= LOGN; j++)
for (int i = 1; i + (1 << j) - 1 <= tot; i++)
f[j][i] = min(f[j-1][i], f[j - 1][i + (1 << (j - 1))]);
for (int i = 1; i <= q; i++)
{
int l = p[rng61() % n + 1], r = p[rng61() % n + 1];
if (l > r) swap(l, r);
int len = 31 - __builtin_clz(r - l + 1);
int d = min(f[len][l], f[len][r - (1 << len) + 1]).second;
ans ^= 1ll * i * d;
}
printf("%lld\n",ans);
return 0;
}
标签:pre,dist,int,rep,cin,DFS,序和树,void,欧拉
来源: https://www.cnblogs.com/guyuLihua/p/16573776.html
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