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bryce_yyds

园龄：2年6个月粉丝：7关注：12

📂做题总结
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2024-02-17 14:09阅读: 5评论: 0推荐: 0

2024 寒假做题总结

P2146 [NOI2015] 软件包管理器

思路分析

树链剖分板子，每次安装时，将 1 到 x 的链变为 1，卸载时，将 x 的子树变为 0。

代码

#include<iostream>

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 1e5 + 10;
int n, q;
struct edge{
	int v, nxt;
}e[N << 1];
int head[N], cnt;
int sz[N], dep[N], son[N], f[N], id[N], top[N], tot;
void add(int u, int v){
	e[++cnt] = (edge){v, head[u]};
	head[u] = cnt;
}
void dfs1(int u, int fa){
	sz[u] = 1, f[u] = fa, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == fa) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[v] > sz[son[u]]) son[u] = v;
	}
}
void dfs2(int u, int t){
	id[u] = ++tot;
	top[u] = t;
	if (son[u]) dfs2(son[u], t);
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == f[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
struct tree{
	int sum, tag;
}t[N << 2];
void pushup(int now){
	t[now].sum = t[now << 1].sum + t[now << 1 | 1].sum;
}
void pushdown(int now, int l, int r){
	if (t[now].tag != -1){
		int mid = (l + r) >> 1;
		t[now << 1].sum = t[now].tag * (mid - l + 1);
		t[now << 1 | 1].sum = t[now].tag * (r - mid);
		t[now << 1].tag = t[now].tag;
		t[now << 1 | 1].tag = t[now].tag;
		t[now].tag = -1;
	}
}
void build(int now, int l, int r){
	t[now].tag = -1, t[now].sum = 0;
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(now << 1, l, mid);
	build(now << 1 | 1, mid + 1, r);
}
void modify(int now, int l, int r, int x, int y, int k){
	if (x <= l && r <= y){
		t[now].sum = k * (r - l + 1), t[now].tag = k;
		return;
	}
	pushdown(now, l, r);
	int mid = (l + r) >> 1;
	if (x <= mid) modify(now << 1, l, mid, x, y, k);
	if (mid + 1 <= y) modify(now << 1 | 1, mid + 1, r, x, y, k);
	pushup(now);
}
void modify_chain(int x, int y, int k){
	int fx = top[x], fy = top[y];
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		modify(1, 1, n, id[fx], id[x], k);
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	modify(1, 1, n, id[x], id[y], k);
}

int main(){
	n = read();
	for (int i = 1; i < n; i++){
		int x = read() + 1;
		add(x, i + 1);
	}
	dfs1(1, 0);
	dfs2(1, 1);
	build(1, 1, n);
	q = read();
	for (int i = 1; i <= q; i++){
		char c[10];
		cin >> c;
		int OP = t[1].sum, x = read() + 1;
		if (c[0] == 'i'){
			modify_chain(1, x, 1);
			cout << t[1].sum - OP << '\n';
		}else{
			modify(1, 1, n, id[x], id[x] + sz[x] - 1, 0);
			cout << OP - t[1].sum << '\n';
		}
	}
	return 0;
}

P2486 [SDOI2011]

思路分析

很好的树链剖分题，在线段树中记录区间最左端颜色和最右端颜色，合并即可。
注意，链与链之间的查询也需要判断合并中间的颜色是否相同。

代码

#include<iostream>
#define ls now << 1
#define rs now << 1 | 1

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 2e5 + 10;
int n, m;
int val[N], sz[N], dep[N], f[N], son[N], id[N], top[N], a[N], tot;
struct edge{
	int v, nxt;
}e[N << 1];
int head[N], cnt;
void add(int u, int v){
	e[++cnt] = (edge){v, head[u]};
	head[u] = cnt;
}
void dfs1(int u, int fa){
	sz[u] = 1, f[u] = fa, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == fa) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v]) son[u] = v;
	}
}
void dfs2(int u, int t){
	top[u] = t;
	id[u] = ++tot;
	a[tot] = val[u];
	if (son[u]) dfs2(son[u], t);
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == f[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
struct tree{
	int lc, rc, sum, l, r, tag;
}t[N << 2];
void pushup(int now){
	t[now].sum = t[ls].sum + t[rs].sum - (t[ls].rc == t[rs].lc);
	t[now].lc = t[ls].lc, t[now].rc = t[rs].rc;
}
void pushdown(int now){
	if (t[now].tag != -1){
		t[ls].lc = t[ls].rc = t[now].tag;
		t[rs].lc = t[rs].rc = t[now].tag;
		t[ls].sum = t[rs].sum = 1;
		t[ls].tag = t[rs].tag = t[now].tag;
		t[now].tag = -1;
	}
}
void build(int now, int l, int r){
	t[now].l = l, t[now].r = r, t[now].sum = t[now].lc = t[now].rc = 0, t[now].tag = -1;
	if (l == r){
		t[now].lc = t[now].rc = a[l];
		t[now].sum = 1;
		return;
	}int mid = (l + r) >> 1;
	build(ls, l, mid);
	build(rs, mid + 1, r);
	pushup(now);
}
void modify(int now, int x, int y, int k){
	int l = t[now].l, r = t[now].r;
	if (x <= l && r <= y){
		t[now].lc = t[now].rc = t[now].tag = k;
		t[now].sum = 1;
		return;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) modify(ls, x, y, k);
	if (mid + 1 <= y) modify(rs, x, y, k);
	pushup(now);
}
void modify_chain(int x, int y, int k){
	int fx = top[x], fy = top[y];
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		modify(1, id[fx], id[x], k);
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	modify(1, id[x], id[y], k);
}
int query(int now, int x, int y){
	int l = t[now].l, r = t[now].r, res = 0;
	if (x <= l && r <= y){
		return t[now].sum;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) res += query(ls, x, y);
	if (mid + 1 <= y) res += query(rs, x, y);
	if(x <= mid && mid + 1 <= y && t[ls].rc == t[rs].lc) res--;
	return res;
}
int check(int now, int x){
	int l = t[now].l, r = t[now].r;
	if (l == r){
		return t[now].lc;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) return check(ls, x);
	else return check(rs, x);
}
int query_chain(int x, int y){
	int fx = top[x], fy = top[y], res = 0;
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		res += query(1, id[fx], id[x]);
		if (check(1, id[fx]) == check(1, id[f[fx]])) res--;
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	res += query(1, id[x], id[y]);
	return res;
}

int main(){
	n = read(), m = read();
	for (int i = 1; i <= n; i++) val[i] = read();
	for (int i = 1; i < n; i++){
		int u = read(), v = read();
		add(u, v), add(v, u);
	}
	dfs1(1, 0);
	dfs2(1, 1);
	build(1, 1, n);
	for (int i = 1; i <= m; i++){
		char opt;
		cin >> opt;
		int x = read(), y = read(), k;
		if (opt == 'C'){
			k = read();
			modify_chain(x, y, k);
		}else{
			cout << query_chain(x, y) << '\n';
		}
	}
	return 0;
}

P7735 [NOI2021] 轻重边

思路分析

将 a 到 b 的路径上每次染上一个新的颜色，则这条边两端的颜色相同，就是重边，否则就是轻边。
注意：每次要先给每个点染上 1 到 n 的颜色，由于染的是点，所以重边的个数是连续相同颜色的点 −1。

代码

#include<iostream>
#include<cstring>
#define ls now << 1
#define rs now << 1 | 1

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 1e5 + 10;
int T, n, m;
int val[N], sz[N], dep[N], f[N], son[N], id[N], top[N], a[N], tot;
struct edge{
	int v, nxt;
}e[N << 1];
int head[N], cnt;
void add(int u, int v){
	e[++cnt] = (edge){v, head[u]};
	head[u] = cnt;
}
void dfs1(int u, int fa){
	sz[u] = 1, f[u] = fa, dep[u] = dep[fa] + 1;
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == fa) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v]) son[u] = v;
	}
}
void dfs2(int u, int t){
	top[u] = t;
	id[u] = ++tot;
	a[tot] = val[u];
	if (son[u]) dfs2(son[u], t);
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == f[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
struct tree{
	int lc, rc, sum, l, r, tag;
}t[N << 2];
void pushup(int now){
	t[now].sum = t[ls].sum + t[rs].sum + (t[ls].rc == t[rs].lc);
	t[now].lc = t[ls].lc, t[now].rc = t[rs].rc;
}
void pushdown(int now){
	if (t[now].tag != -1){
		int l = t[now].l, r = t[now].r, mid = (l + r) >> 1;
		t[ls].lc = t[ls].rc = t[now].tag;
		t[rs].lc = t[rs].rc = t[now].tag;
		t[ls].sum = (mid - l), t[rs].sum = (r - mid - 1);
		t[ls].tag = t[rs].tag = t[now].tag;
		t[now].tag = -1;
	}
}
void build(int now, int l, int r){
	t[now].l = l, t[now].r = r, t[now].sum = 0, t[now].tag = -1;
	if (l == r){
		t[now].lc = t[now].rc = a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(ls, l, mid);
	build(rs, mid + 1, r);
	pushup(now);
}
void modify(int now, int x, int y, int k){
	int l = t[now].l, r = t[now].r;
	if (x <= l && r <= y){
		t[now].lc = t[now].rc = t[now].tag = k;
		t[now].sum = r - l;
		return;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) modify(ls, x, y, k);
	if (mid + 1 <= y) modify(rs, x, y, k);
	pushup(now);
}
void modify_chain(int x, int y, int k){
	int fx = top[x], fy = top[y];
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		modify(1, id[fx], id[x], k);
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	modify(1, id[x], id[y], k);
}
int query(int now, int x, int y){
	int l = t[now].l, r = t[now].r, res = 0;
	if (x <= l && r <= y){
		return t[now].sum;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) res += query(ls, x, y);
	if (mid + 1 <= y) res += query(rs, x, y);
	if(x <= mid && mid + 1 <= y && t[ls].rc == t[rs].lc) res++;
	return res;
}
int check(int now, int x){
	int l = t[now].l, r = t[now].r;
	if (l == r){
		return t[now].lc;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) return check(ls, x);
	else return check(rs, x);
}
int query_chain(int x, int y){
	int fx = top[x], fy = top[y], res = 0;
	while (fx != fy){
		if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy);
		res += query(1, id[fx], id[x]);
		if (check(1, id[fx]) == check(1, id[f[fx]])) res++;
		x = f[fx], fx = top[x];
	}
	if (id[x] > id[y]) swap(x, y);
	res += query(1, id[x], id[y]);
	return res;
}

int main(){
	T = read();
	while (T--){
		memset(t, 0, sizeof t);
		memset(e, 0, sizeof e);
		memset(val, 0, sizeof val);
		memset(sz, 0, sizeof sz);
		memset(dep, 0, sizeof dep);
		memset(f, 0, sizeof f);
		memset(son, 0, sizeof son);
		memset(id, 0, sizeof id);
		memset(a, 0, sizeof a);
		memset(head, 0, sizeof head);
		tot = 0, cnt = 0;
		n = read(), m = read();
		for (int i = 1; i < n; i++){
			int u = read(), v = read();
			add(u, v), add(v, u);
		}
		for (int i = 1; i <= n; i++) val[i] = i;
		dfs1(1, 0);
		dfs2(1, 1);
		build(1, 1, n);
		for (int i = 1, k = n; i <= m; i++){
			int opt = read(), x = read(), y = read();
			if (opt == 1){
				modify_chain(x, y, ++k);
			}else{
				cout << query_chain(x, y) << '\n';
			}
		}
	}
	return 0;
}

P4219 [BJOI2014] 大融合

思路分析

先将这棵树建出来，用一个超级源点连接森林。
并查集维护每个连通块，操作在树上进行。
A 操作，令 x 为较浅的点，rtx 为 x 的祖先，则把 x 到 rtx 的路径上全部加上 size(y)。
Q 询问，令 x 为较浅的点，rtx 为 x 的祖先，则答案为 (size(rtx)−size(y))×size(y)。

代码

#include<iostream>

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 1e5 + 10;
int n, q;
int opt[N], x[N], y[N], fa[N], sz[N], dep[N], f[N], son[N], top[N], id[N], tot;
struct edge{
	int v, nxt;
}e[N << 1];
int head[N], cnt;
void add(int u, int v){
	e[++cnt] = (edge){v, head[u]};
	head[u] = cnt;
}
int find(int x){
	return (x == fa[x] ? x : fa[x] = find(fa[x]));
}
void merge(int x, int y){
	fa[find(y)] = find(x);
}
void dfs1(int u, int rt){
	sz[u] = 1, dep[u] = dep[rt] + 1, f[u] = rt;
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == rt) continue;
		dfs1(v, u);
		sz[u] += sz[v];
		if (sz[son[u]] < sz[v]) son[u] = v;
	}
}
void dfs2(int u, int t){
	id[u] = ++tot;
	top[u] = t;
	if (son[u]) dfs2(son[u], t);
	for (int i = head[u]; i; i = e[i].nxt){
		int v = e[i].v;
		if (v == f[u] || v == son[u]) continue;
		dfs2(v, v);
	}
}
int t[N];
int lowbit(int x){
	return x & (-x);
}
void modify(int x, int k){
	for (int i = x; i <= n; i += lowbit(i)){
		t[i] += k;
	}
}
int query(int x){
	int res = 0;
	for (int i = x; i; i -= lowbit(i)){
		res += t[i];
	}
	return res;
}
void modify_chain(int x, int y, int k){
	int fx = top[x], fy = top[y];
	while (fx != fy){
		if (dep[fx] > dep[fy]) swap(x, y), swap(fx, fy);
		modify(id[fx], k);
		modify(id[x] + 1, -k);
		x = f[fx], fx = top[x];
	}
	if (id[x] < id[y]) swap(x, y);
	modify(id[y], k);
	modify(id[x] + 1, -k);
}

int main(){
	n = read(), q = read();
	for (int i = 1; i <= n; i++) fa[i] = i;
	for (int i = 1; i <= q; i++){
		char c;
		cin >> c;
		if (c == 'A') opt[i] = 1;
		else opt[i] = 2;
		x[i] = read(), y[i] = read();
		if (opt[i] == 1) add(x[i], y[i]), add(y[i], x[i]), merge(x[i], y[i]);
	}
	n++;
	for (int i = 1; i < n; i++){
		if (find(i) == i) add(i, n), add(n, i);
	}
	for (int i = 1; i <= n; i++) fa[i] = i;
	dfs1(n, 0);
	dfs2(n, n);
	modify(1, 1);
	for (int i = 1; i <= q; i++){
		int a = x[i], b = y[i];
		if (f[a] == b) swap(a, b);
		if (opt[i] == 1){
			modify_chain(a, find(a), query(id[b]));
			merge(a, b);
		}else{
			int szx = query(id[find(a)]), szy = query(id[b]);
			cout << (szx - szy) * szy << '\n';
		}
	}
	return 0;
}

P2894 [USACO08FEB] Hotel G

思路分析

线段树难度高一点的题目，要维护区间最大空区间长度和左侧的空区间，右侧的空区间长度。
pushup 中有坑，如果左区间为空，则更新的左侧空区间长度为左区间总长度 + 右区间左侧空区间长度，右区间为空同理。

代码

#include<iostream>
#define ls now << 1
#define rs now << 1 | 1

using namespace std;

inline int read(){register int x = 0, f = 1;register char c = getchar();while (c < '0' || c > '9'){if (c == '-') f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x * f;}
inline void write(int x){if (x < 0) putchar('-'), x = -x;if (x > 9) write(x / 10);putchar(x % 10 + '0');}

const int N = 5e4 + 10;
int n, m;
struct tree{
	int sum, lc, rc, l, r, tag;
}t[N << 2];
void pushup(int now){
	int l = t[now].l, r = t[now].r, mid = (l + r) >> 1;
	t[now].sum = max(max(t[ls].sum, t[rs].sum), t[ls].rc + t[rs].lc);
	if (t[ls].sum == mid - l + 1) t[now].lc = mid - l + 1 + t[rs].lc;
	else t[now].lc = t[ls].lc;
	if (t[rs].sum == r - mid) t[now].rc = r - mid + t[ls].rc;
	else t[now].rc = t[rs].rc;
}
void pushdown(int now){
	int l = t[now].l, r = t[now].r, mid = (l + r) >> 1;
	if (t[now].tag == 0) return;
	if (t[now].tag == 1){
		t[ls].sum = t[ls].lc = t[ls].rc = 0;
		t[rs].sum = t[rs].lc = t[rs].rc = 0;
		t[ls].tag = t[rs].tag = t[now].tag;
		t[now].tag = 0;
	}else{
		t[ls].sum = t[ls].lc = t[ls].rc = (mid - l + 1);
		t[rs].sum = t[rs].lc = t[rs].rc = (r - mid);
		t[ls].tag = t[rs].tag = t[now].tag;
		t[now].tag = 0;
	}
}
void build(int now, int l, int r){
	t[now].l = l, t[now].r = r, t[now].sum = t[now].lc = t[now].rc = (r - l + 1);
	if (l == r) return;
	int mid = (l + r) >> 1;
	build(ls, l, mid);
	build(rs, mid + 1, r);
	pushup(now);
}
void modify(int now, int x, int y){
	int l = t[now].l, r = t[now].r;
	if (x <= l && r <= y){
		t[now].sum = t[now].lc = t[now].rc = 0;
		t[now].tag = 1;
		return;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) modify(ls, x, y);
	if (mid + 1 <= y) modify(rs, x, y);
	pushup(now);
}
void update(int now, int x, int y){
	int l = t[now].l, r = t[now].r;
	if (x <= l && r <= y){
		t[now].sum = t[now].lc = t[now].rc = (r - l + 1);
		t[now].tag = 2;
		return;
	}
	pushdown(now);
	int mid = (l + r) >> 1;
	if (x <= mid) update(ls, x, y);
	if (mid + 1 <= y) update(rs, x, y);
	pushup(now);
}
int query(int now, int x){
	int l = t[now].l, r = t[now].r;
	if (l == r) return l;
	pushdown(now);
	int mid = (l + r) >> 1;
	if (t[ls].sum >= x) return query(ls, x);
	else if (t[ls].rc + t[rs].lc >= x) return mid - t[ls].rc + 1;
	else return query(rs, x);
}

int main(){
	n = read(), m = read();
	build(1, 1, n);
	for (int i = 1; i <= m; i++){
		int opt = read(), x = read(), y;
		if (opt == 1){
			int pos = 0;
			if (t[1].sum >= x){
				pos = query(1, x);
				cout << pos << '\n';
				modify(1, pos, pos + x - 1);
			}else cout << 0 << '\n';
		}else{
			y = read();
			update(1, x, x + y - 1);
		}
	}
	return 0;
}

本文作者：bryce_yyds

本文链接：https://www.cnblogs.com/bryceyyds/p/18017930

版权声明：本作品采用知识共享署名-非商业性使用-禁止演绎 2.5 中国大陆许可协议进行许可。

posted @   bryce_yyds  阅读(5)  评论(0)  编辑  收藏  举报
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